Ordinary Di erential Equations. Session 7roquesol/Math_308_Fall_2017_Session_7.pdfIntroduction...

Systems of First Order Linear EquationsSystems of First Order Linear Equations II

Ordinary Differential Equations. Session 7

Dr. Marco A Roque Sol

11/09/2017

Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

IntroductionSystems of Linear Algebraic Equations; Linear Independence, Eigenvalues, EigenvectorsSystems of Linear Algebraic Equations; Linear Independence, Eigenvalues, EigenvectorsBasic Theory of Systems of First Order Linear Equations

Review of Matrices

Let us start by solving an m × n system of linear equations

a11x1 + a12x2 + . . .+ a1nxn = b1a21x1 + a22x2 + . . .+ a2nxn = b2

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients,

b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, and

x ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns.

In this way, we can introduce new arrays ofnumbers to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

where aij are given coefficients, b′ms are given right-hand side, andx ′ns are the unknowns. In this way, we can introduce new arrays ofnumbers

to study the linear system

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

Review of Matrices

In this way, we have the following

Definition

An m× n matrix A , is an array of complex numbers ( m-rows andn-columns ),denoted by

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

In this context, an element in the i-row and j-column is of thematrix A denoted by aij .

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

An m× n matrix A ,

is an array of complex numbers ( m-rows andn-columns ),denoted by

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

An m× n matrix A , is an array of complex numbers ( m-rows andn-columns ),

denoted by

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

In this context,

an element in the i-row and j-column is of thematrix A denoted by aij .

Review of Matrices

Definition

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n

Review of Matrices

Associated with any m × n A matrix, we have the followingmatrices:a) Transpose

Is the ( n ×m ) matrix, denoted by AT , and defined by

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

b) Complex Conjugate

Is the ( m × n ) matrix, denoted by A, and defined by

Review of Matrices

Associated with any

m × n A matrix, we have the followingmatrices:a) Transpose

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

Associated with any m × n A matrix,

we have the followingmatrices:a) Transpose

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

Associated with any m × n A matrix, we have the followingmatrices:

a) Transpose

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

Is the ( n ×m ) matrix, denoted by AT , and

defined by

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Is the ( m × n ) matrix, denoted by A, and

defined by

Review of Matrices

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(aTij

= (aji )m×n

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

= (aij)m×n = (aij)m×n

c) Adjoint

Is the ( m × n ) matrix, denoted by A∗ = AT

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

=(a∗ij)n×m = (aji )m×n

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

c) Adjoint

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

c) Adjoint

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

c) Adjoint

defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

c) Adjoint

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

Review of Matrices

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

c) Adjoint

, and defined by

A∗ =

a11 a21 . . . am1

a12 a22 . . . am2...

a1n a2n . . . amn

Review of Matrices

Basic Matrix Operations

Let A = (aij)m×n and B = (bij)m×n be two matrices, then wedefine

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

A± B = (aij ± bij)m×ne) Scalar Multiplication

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

A± B =

(aij ± bij)m×ne) Scalar Multiplication

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

A± B = (aij ± bij)m×n

e) Scalar Multiplication

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

(raij)m×n

Review of Matrices

1) A = B ⇐⇒

aij = bij ; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Addtion

rA = (raij)m×nDr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Review of Matrices

Matrix Multiplication

Let A and B, m × p and p × n matrices respectively

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

AB = (cij)m×n

p∑k=1

aikbkj

(AB)ij = cij =

. . . . . .. . . . . .ai1 ai2 . . . ain

. . ....

. . . b1j . . .

. . . b2j . . .

. . .... . . .

. . . bnj . . .

Review of Matrices

In general, when AB is defined, not necessarily BA is also defined,but even in that case, we have in general

AB 6= BA

Example 7.1

Let A and B the matrices defined by

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Find A + B, A− B, 3A AB, BA

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

In general, when AB is defined, not necessarily BA is also defined,

but even in that case, we have in general

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

AB 6= BA

Example 7.1

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Solution

A + B =

3 −1 01 1 −14 0 2

A− B =

−1 −3 2−1 3 −10 2 0

6 3 −3−3 3 06 −3 3

1 −2 10 2 −12 1 1

2 1 −11 −1 02 −1 1

2 2 00 −1 −17 0 −1

2 1 −11 −1 02 −1 1

1 −2 10 2 −12 1 1

0 −3 01 0 24 −5 4

Review of Matrices

Example 7.2

Let C and D the matrices defined by

2 11 −12 −1

(1 −2 10 2 −1

)Find CD and DC.Solution

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

Find CD and DC.Solution

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

)Find CD and DC.

Solution

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.2

2 11 −12 −1

(1 −2 10 2 −1

2 11 −12 −1

(1 −2 10 2 −1

2 −2 11 −4 22 −6 3

(1 −2 10 2 −1

) 2 11 −12 −1

(2 20 −1

)6= CD

Review of Matrices

Example 7.3

Using matrix operations rewrite the linear system

......

am1x1 + am2x2 + . . .+ amnxn = bm

in terms of matrices.

Solution

Starting with the system

Review of Matrices

Example 7.3

......

am1x1 + am2x2 + . . .+ amnxn = bm

Solution

Review of Matrices

Example 7.3

......

am1x1 + am2x2 + . . .+ amnxn = bm

Solution

Review of Matrices

Example 7.3

......

am1x1 + am2x2 + . . .+ amnxn = bm

Solution

Review of Matrices

Example 7.3

......

am1x1 + am2x2 + . . .+ amnxn = bm

Solution

Review of Matrices

Example 7.3

......

am1x1 + am2x2 + . . .+ amnxn = bm

Solution

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

......

am1x1 + am2x2 + . . .+ amnxn = bm

and choosing

a11 a12 . . . a1na21 a22 . . . a2n

...am1 am2 . . . amn

x1x2...xm

b1b2...bm

we get

Review of Matrices

a11x1 + a12x2 + . . .+ a1nxna21x1 + a22x2 + . . .+ a2nxn

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

Types of Matrices An m × n matrix A = (aij)m×n is a

1) Zero matrix if aij = 0; i = 1, 2, ...,m, j = 1, 2, ..., n

2) Square Matrix if m = n.

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒

AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

Types of Matrices

An m × n matrix A = (aij)m×n is a

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

Review of Matrices

...am1x1 + am2x2 + . . .+ amnxn

b1b2...bm

= B =⇒ AX = B

2 −2 11 −4 22 −6 3

(3 75 −4

)Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Review of Matrices

3) Identity matrix (n × n) (I) if aij = δij where δij =

{1 i = j0 i 6= j

A = I =

1. . .

4) Symetric Matrix (n × n) if AT = A or aij = aji ;

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

3) Identity matrix (n × n)

(I) if aij = δij where δij =

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

3) Identity matrix (n × n) (I)

if aij = δij where δij =

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

3) Identity matrix (n × n) (I) if aij = δij

where δij =

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

{1 i = j0 i 6= j

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

{1 i = j0 i 6= j

A = I =

1. . .

4) Symetric Matrix (n × n)

if AT = A or aij = aji ;i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

{1 i = j0 i 6= j

A = I =

1. . .

i = 1, 2, ...,m, j = 1, 2, ..., n

Review of Matrices

5) Triangular Matrix (n × n)

5a) Upper Triangular Matrix(U) if uij = 0, i > j

a11 · · · · · · a1n

a22. . .

5b) Lower Triangular Matrix(L) if lij = 0, i < j

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

5a) Upper Triangular Matrix

(U) if uij = 0, i > j

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

5a) Upper Triangular Matrix(U)

if uij = 0, i > j

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

5b) Lower Triangular Matrix

(L) if lij = 0, i < j

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

5b) Lower Triangular Matrix(L)

if lij = 0, i < j

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

a11 · · · · · · a1n

a22. . .

a22 0. . .

... · · · · · · ann

Review of Matrices

6) Diagonal Matrix (n × n) (D) if aij = dij where dij = Diδij

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

7) Invertible Matrix (n × n) If A is a square matrix (n × n) andthere exists an n × n matrix B such that

AB = BA = I

The matrix B is denoted by A−1 and is called the Inverse Matrixand A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

6) Diagonal Matrix (n × n)

(D) if aij = dij where dij = Diδij

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

6) Diagonal Matrix (n × n) (D)

if aij = dij where dij = Diδij

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

6) Diagonal Matrix (n × n) (D) if aij = dij where

dij = Diδij

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

7) Invertible Matrix (n × n)

If A is a square matrix (n × n) andthere exists an n × n matrix B such that

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

7) Invertible Matrix (n × n) If A is a square matrix (n × n) and

there exists an n × n matrix B such that

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA =

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

The matrix B

is denoted by A−1 and is called the Inverse Matrixand A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

The matrix B is denoted by

A−1 and is called the Inverse Matrixand A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

The matrix B is denoted by A−1 and

is called the Inverse Matrixand A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

The matrix B is denoted by A−1 and is called the Inverse Matrixand

A is called invertible or nonsingular matrix. Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

The matrix B is denoted by A−1 and is called the Inverse Matrixand A is called invertible or nonsingular matrix.

Matrices that donot have an inverse are called it singular or noninvertible.

Review of Matrices

a11 · · · · · ·

a22 00 . . .

... · · · · · · ann

AB = BA = I

Review of Matrices

A−1 is the notation for the inverse of A, but keep in mind that

A−1 6= 1

There are various ways to compute A−1 from A, assuming that itexists. One way is the cofactor expansion .

Associated with each element aij of a given matrix is the minor Mij

from, which is the determinant of the matrix obtained by deletingthe ith row and jth column of the original matrix that is, the rowand column containing aij . Also associated with each element aij isthe cofactor Cij defined by the equation

Review of Matrices

A−1 6= 1

Review of Matrices

A−1 is the notation for the inverse of A,

but keep in mind that

A−1 6= 1

Review of Matrices

A−1 6= 1

Review of Matrices

A−1 6= 1

Review of Matrices

A−1 6= 1

There are various ways to compute A−1 from A, assuming that itexists.

One way is the cofactor expansion .

Review of Matrices

A−1 6= 1

Review of Matrices

A−1 6= 1

Associated with each element aij of a given matrix

is the minor Mij

Review of Matrices

A−1 6= 1

which is the determinant of the matrix obtained by deletingthe ith row and jth column of the original matrix that is, the rowand column containing aij . Also associated with each element aij isthe cofactor Cij defined by the equation

Review of Matrices

A−1 6= 1

from, which is the determinant of the matrix obtained by deletingthe ith row and

jth column of the original matrix that is, the rowand column containing aij . Also associated with each element aij isthe cofactor Cij defined by the equation

Review of Matrices

A−1 6= 1

from, which is the determinant of the matrix obtained by deletingthe ith row and jth column of the original matrix

that is, the rowand column containing aij . Also associated with each element aij isthe cofactor Cij defined by the equation

Review of Matrices

A−1 6= 1

from, which is the determinant of the matrix obtained by deletingthe ith row and jth column of the original matrix that is, the rowand column containing aij .

Also associated with each element aij isthe cofactor Cij defined by the equation

Review of Matrices

A−1 6= 1

Review of Matrices

Cij = (−1)nMij

If B = A−1, then it can be shown that the general element bij isgiven by

bij =Cij

det(A)

In general the use of the above equation is not an efficient way tocalculate A−1, instead we can use elementary row operations.There are three such operations:

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

Review of Matrices

Cij = (−1)nMij

If B = A−1, then it can be shown that

the general element bij isgiven by

bij =Cij

det(A)

Review of Matrices

Cij = (−1)nMij

If B = A−1, then it can be shown that the general element bij

isgiven by

bij =Cij

det(A)

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

In general the use of the above equation is not an efficient way tocalculate A−1, instead

we can use elementary row operations.There are three such operations:

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

In general the use of the above equation is not an efficient way tocalculate A−1, instead we can use elementary row operations.

There are three such operations:

Review of Matrices

Cij = (−1)nMij

bij =Cij

det(A)

Review of Matrices

1. Interchange of two rows.

2. Multiplication of a row by a nonzero scalar.

3. Addition of any multiple of one row to another row.

The transformation of a matrix by a sequence of elementary rowoperations is referred to as row reduction or Gaussian elimination.Starting with the matrix A we build the m × 2n Augment Matrix

[A | I]

and using elementary row operations we tranforme it into[I | A−1

Review of Matrices

[A | I]

Review of Matrices

[A | I]

Review of Matrices

[A | I]

Review of Matrices

The transformation of a matrix by

a sequence of elementary rowoperations is referred to as row reduction or Gaussian elimination.Starting with the matrix A we build the m × 2n Augment Matrix

[A | I]

Review of Matrices

The transformation of a matrix by a sequence of elementary rowoperations

is referred to as row reduction or Gaussian elimination.Starting with the matrix A we build the m × 2n Augment Matrix

[A | I]

Review of Matrices

The transformation of a matrix by a sequence of elementary rowoperations is referred to as row reduction or Gaussian elimination.

Starting with the matrix A we build the m × 2n Augment Matrix

[A | I]

Review of Matrices

The transformation of a matrix by a sequence of elementary rowoperations is referred to as row reduction or Gaussian elimination.Starting with the matrix A

we build the m × 2n Augment Matrix

[A | I]

Review of Matrices

[A | I]

Review of Matrices

[A | I]

Review of Matrices

[A | I]

and using elementary row operations we tranforme it into

[I | A−1

Review of Matrices

[A | I]

]Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Review of Matrices

Example 7.4

Find the inverse of

1 −1 −13 −1 22 2 3

Solution

First of all, let’s build the augmented matrix

1 −1 −1

∣∣∣ 1 0 0

3 −1 2∣∣∣ 0 1 0

2 2 3∣∣∣ 0 0 1

Review of Matrices

Example 7.4

Find the inverse of

1 −1 −13 −1 22 2 3

Solution

1 −1 −1

∣∣∣ 1 0 0

3 −1 2∣∣∣ 0 1 0

2 2 3∣∣∣ 0 0 1

Review of Matrices

Example 7.4

Find the inverse of

1 −1 −13 −1 22 2 3

Solution

1 −1 −1

∣∣∣ 1 0 0

3 −1 2∣∣∣ 0 1 0

2 2 3∣∣∣ 0 0 1

Review of Matrices

Example 7.4

Find the inverse of

1 −1 −13 −1 22 2 3

Solution

1 −1 −1

∣∣∣ 1 0 0

3 −1 2∣∣∣ 0 1 0

2 2 3∣∣∣ 0 0 1

Review of Matrices

Example 7.4

Find the inverse of

1 −1 −13 −1 22 2 3

Solution

1 −1 −1

∣∣∣ 1 0 0

3 −1 2∣∣∣ 0 1 0

2 2 3∣∣∣ 0 0 1

Review of Matrices

(a) Obtain zeros in the off-diagonal positions in the first column byadding (−3) times the first row to the second row and adding(−2) times the first row to the third row.

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

(b) Obtain a 1 in the diagonal position in the second column bymultiplying the second row by 1/2 .

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

(a) Obtain zeros in the off-diagonal positions in the first column byadding

(−3) times the first row to the second row and adding(−2) times the first row to the third row.

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

(a) Obtain zeros in the off-diagonal positions in the first column byadding (−3) times the first row

to the second row and adding(−2) times the first row to the third row.

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

(a) Obtain zeros in the off-diagonal positions in the first column byadding (−3) times the first row to the second row and

adding(−2) times the first row to the third row.

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

(a) Obtain zeros in the off-diagonal positions in the first column byadding (−3) times the first row to the second row and adding(−2) times the first row

to the third row.

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

(b) Obtain a 1 in the diagonal position in the second column

bymultiplying the second row by 1/2 .

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

1 −1 −1

∣∣∣ 1 0 0

0 2 5∣∣∣ −3 1 0

0 4 5∣∣∣ −2 0 1

1 −1 −1

∣∣∣ 1 0 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 4 5∣∣∣ −2 0 1

Review of Matrices

(c) Obtain zeros in the off-diagonal positions in the second columnby adding the second row to the first row and adding (−4) timesthe second row to the third row

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

(d) Obtain a 1 in the diagonal position in the third column bymultiplying the third row by (−1/5).

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

(c) Obtain zeros in the off-diagonal positions in the second columnby adding

the second row to the first row and adding (−4) timesthe second row to the third row

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

(c) Obtain zeros in the off-diagonal positions in the second columnby adding the second row

to the first row and adding (−4) timesthe second row to the third row

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

(c) Obtain zeros in the off-diagonal positions in the second columnby adding the second row to the first row and

adding (−4) timesthe second row to the third row

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

(c) Obtain zeros in the off-diagonal positions in the second columnby adding the second row to the first row and adding (−4) timesthe second row

to the third row

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

(d) Obtain a 1 in the diagonal position in the third column

bymultiplying the third row by (−1/5).

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

(d) Obtain a 1 in the diagonal position in the third column bymultiplying the third row

by (−1/5).

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 −5∣∣∣ 4 −2 1

1 0 3/2

∣∣∣ −1/2 1/2 0

0 1 5/2∣∣∣ −3/2 1/2 0

0 0 1∣∣∣ −4/5 2/5 −1/5

Review of Matrices

(e) Obtain zeros in the off-diagonal positions in the third columnby adding (−3/2) times the third row to the first row and adding(−5/2) times the third row to the second row.

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

so, the inverse matrix A−1 is given by

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

(e) Obtain zeros in the off-diagonal positions in the third column

by adding (−3/2) times the third row to the first row and adding(−5/2) times the third row to the second row.

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

(e) Obtain zeros in the off-diagonal positions in the third columnby adding (−3/2) times the third row

to the first row and adding(−5/2) times the third row to the second row.

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

(e) Obtain zeros in the off-diagonal positions in the third columnby adding (−3/2) times the third row to the first row

and adding(−5/2) times the third row to the second row.

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

(e) Obtain zeros in the off-diagonal positions in the third columnby adding (−3/2) times the third row to the first row and adding(−5/2) times the third row

to the second row.

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

∣∣∣ 7/10 −1/10 3/10

0 1 0∣∣∣ 1/2 −1/2 1/2

0 0 1∣∣∣ −4/5 2/5 −1/5

A−1 =

7/10 −1/10 3/101/2 −1/2 1/2−4/5 2/5 −1/5

7 −1 35 −5 5−8 4 −2

Review of Matrices

Matrix Functions .We sometimes need to consider vectors or matrices whoseelements are functions of a real variable t. In that case, we write

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

The matrix A(t) is said to be continuous at t = t0 or on aninterval α < t < β if each element of A(t) is a continuous functionat the given point or on the given interval.

Review of Matrices

Matrix Functions .

We sometimes need to consider vectors or matrices whoseelements are functions of a real variable t. In that case, we write

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

Matrix Functions .We sometimes need to consider vectors or

matrices whoseelements are functions of a real variable t. In that case, we write

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

Matrix Functions .We sometimes need to consider vectors or matrices whoseelements are

functions of a real variable t. In that case, we write

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

Matrix Functions .We sometimes need to consider vectors or matrices whoseelements are functions of a real variable t.

In that case, we write

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

The matrix A(t) is said to be continuous at t = t0 or

on aninterval α < t < β if each element of A(t) is a continuous functionat the given point or on the given interval.

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

The matrix A(t) is said to be continuous at t = t0 or on aninterval α < t < β if

each element of A(t) is a continuous functionat the given point or on the given interval.

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

The matrix A(t) is said to be continuous at t = t0 or on aninterval α < t < β if each element of A(t) is a continuous function

at the given point or on the given interval.

Review of Matrices

X(t) =

x1(t)x2(t)

...xn(t)

= A(t) =

a11(t) · · · a1n(t)...

...am1(t) amn(t)

respectively.

Continuity

Review of Matrices

Differentiability

Similarly, A(t) is said to be differentiable if each of its elements isdifferentiable, and its derivative dA(t)/dt is defined by

(daij(t)

that is, each element of dA(t)/dt is the derivative of thecorresponding element of A(t).

Integrability

In the same way, the integral of a matrix function is defined as∫ b

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

Similarly, A(t) is said to be differentiable if

each of its elements isdifferentiable, and its derivative dA(t)/dt is defined by

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

Similarly, A(t) is said to be differentiable if each of its elements isdifferentiable, and

its derivative dA(t)/dt is defined by

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

that is, each element of dA(t)/dt

is the derivative of thecorresponding element of A(t).

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

In the same way, the integral of a matrix function is defined as

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Differentiability

(daij(t)

Integrability

aA(t)dt =

(∫ b

aaij(t)dt

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

0 tdt∫ π0 cos(t)dt

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

Find A′(t) and∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Review of Matrices

Example 7.5

Consider the matrix

A(t) =

(sin(t) 1t cos(t)

)Find A′(t) and

∫ π0 A(t)dt.

Solution

A′(t) =

(cos(t) 0

1 −sin(t)

)∫ π

0A(t)dt =

(∫ π0 sin(t)dt

∫ π0 1dt∫ π

π2/2 0

Systems of Linear Algebraic Equations; LinearIndependence, Eigenvalues, Eigenvectors

Systems of Linear Algebraic Equations . A set of n simultaneouslinear algebraic equations in n variables

......

an1x1 + an2x2 + . . .+ annxn = bnn

can be written as

AX = b

Systems of Linear Algebraic Equations .

A set of n simultaneouslinear algebraic equations in n variables

......

can be written as

AX = b

......

can be written as

AX = b

......

can be written as

AX = b

......

can be written as

AX = b

......

can be written as

AX = b

If b = 0, the system is said to be homogeneous; otherwise, it isnonhomogeneous.

If the matrix A is invertible,hence A−1 exists, and therefore wehave

X = A−1b

In particular, the homogeneous problem AX = b, corresponding tob = 0, has only the trivial solution 0.

If b = 0, the system is said to be homogeneous;

otherwise, it isnonhomogeneous.

X = A−1b

If the matrix A is invertible,

hence A−1 exists, and therefore wehave

X = A−1b

If the matrix A is invertible,hence A−1 exists, and

therefore wehave

X = A−1b

In particular,

the homogeneous problem AX = b, corresponding tob = 0, has only the trivial solution 0.

X = A−1b

In particular, the homogeneous problem AX = b,

corresponding tob = 0, has only the trivial solution 0.

X = A−1b

In particular, the homogeneous problem AX = b, corresponding tob = 0,

has only the trivial solution 0.

X = A−1b

On the other hand, if A is singular, A−1 does not exist, so thehomogeneous system

AX = 0

has (infinitely many) nonzero solutions in addition to the trivialsolution.

Solving a Linear System

For solving particular systems, we can form the augmented matrix

On the other hand,

if A is singular, A−1 does not exist, so thehomogeneous system

AX = 0

On the other hand, if A is singular,

A−1 does not exist, so thehomogeneous system

AX = 0

On the other hand, if A is singular, A−1 does not exist,

so thehomogeneous system

AX = 0

For solving particular systems,

we can form the augmented matrix

AX = 0

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

We now perform row operations on the augmented matrix so as totransform A into an upper triangular matrix.

[U|b̄]

Once this is done, it is easy to see whether the system hassolutions, and to find them if it does.

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

We now perform row operations on the augmented matrix

so as totransform A into an upper triangular matrix.

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

We now perform row operations on the augmented matrix so as totransform A

into an upper triangular matrix.

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

Once this is done,

it is easy to see whether the system hassolutions, and to find them if it does.

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

Once this is done, it is easy to see whether the system hassolutions, and

to find them if it does.

[A|b] =

a11 a12 . . . a1n

∣∣∣ b1

a21 a22 . . . a2n

∣∣∣ b2...

∣∣∣ ...

an1 an2 . . . ann

∣∣∣ bn

[U|b̄]

Example 7.6

Solve the system of equations

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

The augmented matrix for the system is1 −2 3

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

The augmented matrix for the system is

1 −2 3

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

Example 7.6

x1 − 2x2 + 3x3 = 7−x1 + x2 − 2x3 = −52x1 − x2 − x3 = 4

Solution

∣∣∣ 7

−1 1 −2∣∣∣ −5

2 −1 −1∣∣∣ 4

We now perform row operations on the augmented matrix with aview to introducing zeros in the lower left part of the matrix.

(a) Add the first row to the second row, and add (−2) times thefirst row to the third row.

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

We now perform row operations on the augmented matrix

with aview to introducing zeros in the lower left part of the matrix.

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

(a) Add the first row to the second row,

and add (−2) times thefirst row to the third row.

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

(a) Add the first row to the second row, and add (−2) times thefirst row

to the third row.

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

1 −2 3

∣∣∣ 7

0 −1 1∣∣∣ 2

0 −3 −7∣∣∣ −10

(b) Multiply the second row by −1.1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

(c) Add (−3) times the second row to the third row.

1 −2 3∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

(b) Multiply the second row by −1.

1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

1 −2 3∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

(c) Add (−3) times the second row to the third row.1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

(c) Add (−3) times the second row

to the third row.1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

∣∣∣ 7

0 1 −1∣∣∣ −2

0 3 −7∣∣∣ −10

1 −2 3∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 −4∣∣∣ −4

(d) Divide the third row by −4.1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 1∣∣∣ 1

The matrix obtained in this manner corresponds to the system ofequations

x1 − 2x2 + 3x3 = 7x2 − x3 = −2

x3 = 1

(d) Divide the third row by −4.

1 −2 3

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 1∣∣∣ 1

x1 − 2x2 + 3x3 = 7x2 − x3 = −2

x3 = 1

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 1∣∣∣ 1

x1 − 2x2 + 3x3 = 7x2 − x3 = −2

x3 = 1

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 1∣∣∣ 1

x1 − 2x2 + 3x3 = 7x2 − x3 = −2

x3 = 1

∣∣∣ 7

0 1 −1∣∣∣ −2

0 0 1∣∣∣ 1

x1 − 2x2 + 3x3 = 7x2 − x3 = −2

x3 = 1

From the last of equations we have

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

Now, since the solution is unique, we conclude that the coefficientmatrix is nonsingular.

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2,

x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 =

− 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1,

x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 =

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

Now, since the solution is unique,

we conclude that the coefficientmatrix is nonsingular.

x3 = 2, x2 = −2 + x3 = − 1, x3 = 7 + 2x2 − 2x3 = 2

Thus, we obtain

2− 1

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

for various values of b1, b2, and b3

Solution

By performing steps (a), (b), and (c) as in Example 7.6, wetransform the matrix into

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

By performing steps (a), (b), and (c) as in Example 7.6,

wetransform the matrix into

Example 7.7

x1 − 2x2 + 3x3 = b1−x1 + x2 − 2x3 = b22x1 − x2 − 3x3 = b3

Solution

1 −2 3

∣∣∣ b1

0 1 −1∣∣∣ −b1 − b2

0 0 0∣∣∣ b1 + 3b2 + b3

The equation corresponding to the third row is

b1 + 3b2 + b3 = 0

1 −2 3

∣∣∣ b1

0 1 −1∣∣∣ −b1 − b2

0 0 0∣∣∣ b1 + 3b2 + b3

b1 + 3b2 + b3 = 0

1 −2 3

∣∣∣ b1

0 1 −1∣∣∣ −b1 − b2

0 0 0∣∣∣ b1 + 3b2 + b3

b1 + 3b2 + b3 = 0

1 −2 3

∣∣∣ b1

0 1 −1∣∣∣ −b1 − b2

0 0 0∣∣∣ b1 + 3b2 + b3

b1 + 3b2 + b3 = 0

thus the system has no solution unless the above condition issatisfied by b1, b2, and b3.

b1 = −3b2 − b3

Assuming that the condition is satisfied1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

thus the system has no solution unless

the above condition issatisfied by b1, b2, and b3.

b1 = −3b2 − b3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

b1 = −3b2 − b3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

b1 = −3b2 − b3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

b1 = −3b2 − b3

Assuming that the condition is satisfied

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

b1 = −3b2 − b3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ −(−3b2 − b3)− b2

0 0 0∣∣∣ (−3b2 − b3) + 3b2 + b3

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ 2b2 + b3

0 0 0∣∣∣ 0

Add (2) times the second row to the first row.

1 0 1∣∣∣ −3b2 − b3

0 1 −1∣∣∣ b2 + b3

0 0 0∣∣∣ 0

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ 2b2 + b3

0 0 0∣∣∣ 0

Add (2) times the second row to the first row.1 0 1

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ b2 + b3

0 0 0∣∣∣ 0

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ 2b2 + b3

0 0 0∣∣∣ 0

Add (2) times the second row

to the first row.1 0 1

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ b2 + b3

0 0 0∣∣∣ 0

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ 2b2 + b3

0 0 0∣∣∣ 0

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ b2 + b3

0 0 0∣∣∣ 0

1 −2 3

∣∣∣ −3b2 − b3

0 1 −1∣∣∣ 2b2 + b3

0 0 0∣∣∣ 0

1 0 1∣∣∣ −3b2 − b3

0 1 −1∣∣∣ b2 + b3

0 0 0∣∣∣ 0

Thus, we have two equations and one unknown,so one of thevariables let’s say x3, is equal to a parameter α, obtaining thesystem

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

Thus, we have two equations and one unknown,

so one of thevariables let’s say x3, is equal to a parameter α, obtaining thesystem

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

Thus, we have two equations and one unknown,so one of thevariables let’s say x3,

is equal to a parameter α, obtaining thesystem

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

x1 + α = −3b2 − b3x2 − α = b2 + b3

Hence, we obtain

x1 = −α− 3b2 − b3; x2 = α + b2 + b3

−α− 3b2 − b3α + b2 + b3

−111

−3b2 − b3b2 + b3

Linear Dependence and Independence .

A set of k vectors x(1), ..., x(k) is said to be linearly dependent ifthere exists a set of real or complex numbers c1, ...., ck , at leastone of which is nonzero, such that

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k) is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

A set of k vectors x(1), ..., x(k)

is said to be linearly dependent ifthere exists a set of real or complex numbers c1, ...., ck , at leastone of which is nonzero, such that

c1x(1) + ...+ ckx(k) = 0

A set of k vectors x(1), ..., x(k) is said to be linearly dependent

ifthere exists a set of real or complex numbers c1, ...., ck , at leastone of which is nonzero, such that

c1x(1) + ...+ ckx(k) = 0

A set of k vectors x(1), ..., x(k) is said to be linearly dependent ifthere exists a set of real or complex numbers c1, ...., ck ,

at leastone of which is nonzero, such that

c1x(1) + ...+ ckx(k) = 0

A set of k vectors x(1), ..., x(k) is said to be linearly dependent ifthere exists a set of real or complex numbers c1, ...., ck , at leastone of which is nonzero,

such that

c1x(1) + ...+ ckx(k) = 0

On the other hand,

if the only set c1, ..., ck for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k) is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck

for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k) is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck for which the aboveequation is satisfied is

c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k) is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then

the set ofvectors x(1), ..., x(k) is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k)

is called linearly independent.

c1x(1) + ...+ ckx(k) = 0

Consider now a set of n vectors, each of which has n components ,

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

the above equation can be written as .

x11c1 + x12c2 + . . .+ x1ncnx21c1 + x22c2 + . . .+ x2ncn

......

xn1c1 + xn2c2 + . . .+ xnncn

Consider now a set of n vectors,

each of which has n components ,

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

· · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · ·

x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

the above equation can be written as .x11c1 + x12c2 + . . .+ x1ncnx21c1 + x22c2 + . . .+ x2ncn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

......

xn1c1 + xn2c2 + . . .+ xnncn

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then the only solution of is c = 0,but if X is singular (X−1 does not exist) there are nonzerosolutions.

Example 7.8

Determine wether the vectors are linearly indepent or not

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

If X is nonsingular

(X−1 exists), then the only solution of is c = 0,but if X is singular (X−1 does not exist) there are nonzerosolutions.

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then

the only solution of is c = 0,but if X is singular (X−1 does not exist) there are nonzerosolutions.

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then the only solution of is c = 0,

but if X is singular (X−1 does not exist) there are nonzerosolutions.

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then the only solution of is c = 0,but if X is singular

(X−1 does not exist) there are nonzerosolutions.

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then the only solution of is c = 0,but if X is singular (X−1 does not exist)

there are nonzerosolutions.

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

x(2) =

; x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

x(3) =

− 41

− 11

or equivalently

Xc = 0

Example 7.8

x(1) =

; x(2) =

; x(3) =

− 41

− 11

Solution

To determine whether x (1), x (2), and x (3) are linearly dependent,we seek constants c1, c2, and c3 such that

c1x(1) + c2x(2) + c3x(3) = 0

written in the matrix form 1 2 42 1 1−1 3 −11

c1c2c3

Solution

c1x(1) + c2x(2) + c3x(3) = 0

c1c2c3

Solution

To determine whether x (1), x (2), and x (3) are linearly dependent,

we seek constants c1, c2, and c3 such that

c1x(1) + c2x(2) + c3x(3) = 0

c1c2c3

Solution

c1x(1) + c2x(2) + c3x(3) = 0

c1c2c3

Solution

c1x(1) + c2x(2) + c3x(3) = 0

c1c2c3

Solution

c1x(1) + c2x(2) + c3x(3) = 0

written in the matrix form

1 2 42 1 1−1 3 −11

c1c2c3

Solution

c1x(1) + c2x(2) + c3x(3) = 0

c1c2c3

Using elementary row operations on the augmented matrix1 2 −4

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

(a) Add (−2) times the first row to the second row, and add thefirst row to the third row.

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

Using elementary row operations on the augmented matrix

1 2 −4

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

(a) Add (−2) times the first row

to the second row, and add thefirst row to the third row.

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

(a) Add (−2) times the first row to the second row, and

add thefirst row to the third row.

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

(a) Add (−2) times the first row to the second row, and add thefirst row

to the third row.1 2 −4

∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

1 2 −4

∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0

(b) Divide the second row by (−3), then add (−5) times thesecond row to the third row.

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

Thus we obtain the equivalent system(

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

(b) Divide the second row by (−3),

then add (−5) times thesecond row to the third row.

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

(b) Divide the second row by (−3), then add (−5) times thesecond row

to the third row.1 2 −4

∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

1 2 −4

∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

Thus we obtain the equivalent system(c1 + 2c2 − 4c3 = 0

c2 − 3c3 = 0

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

Thus we obtain the equivalent system

(c1 + 2c2 − 4c3 = 0

c2 − 3c3 = 0

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

Hence, we have 2 equations in 3 unknowns, so one of them, let’ssay c3 will be a free parameter (real number) α and the solution ofthe system is

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

Hence, there are infinitely solutions and the set of vectors islinearly dependent.

Hence, we have 2 equations in 3 unknowns, so

one of them, let’ssay c3 will be a free parameter (real number) α and the solution ofthe system is

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

Hence, we have 2 equations in 3 unknowns, so one of them, let’ssay c3 will be a free parameter (real number) α and

the solution ofthe system is

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α;

c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α;

c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

Hence, there are infinitely solutions and

the set of vectors islinearly dependent.

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

− 2α3αα

− 231

Determinants

Associated to every n × n matrix A there is a real number calledthe determinant of A denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

Associated to every n × n matrix A

there is a real number calledthe determinant of A denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

Associated to every n × n matrix A there is a real number calledthe determinant of A

denoted by |A| or det(A) and definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

Associated to every n × n matrix A there is a real number calledthe determinant of A denoted by |A| or det(A) and

definedinductivly as follows

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11

|A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ =

a11a22 − a12a21

Determinants

n = 1 A = a11 |A| = a11

n = 2 A =

(a11 a12a21 a22

)|A| =

∣∣∣∣a11 a12a21 a22

∣∣∣∣ = a11a22 − a12a21

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12 =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣−

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 +

a13a21a32 − a31a22a13 − a32a23a11 − a33a21a12 =

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 + a13a21a32 −

a31a22a13 − a32a23a11 − a33a21a12 =

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 + a13a21a32 − a31a22a13 −

a32a23a11 − a33a21a12 =

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a12a23a31 + a13a21a32 − a31a22a13 − a32a23a11 −

a33a21a12 =

n = 3 A =

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ =

∣∣∣∣a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣a11 a12a31 a32

∣∣∣∣ = a11a22a33+

a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−a32a23a11 − a33a21a12

∣∣∣∣∣∣∣∣a11 a12 a13 a11 a12a21 a22 a23 a21 a22a31 a32 a33 a31 a32

− − − + + +

∣∣∣∣∣∣∣∣Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

(−1)1+ja1jM1j

or using any fix row i

|A| =n∑

(−1)i+jaijMij

or using any fix column j

|A| =n∑

(−1)i+jaijMij

a11a22a33 + a12a23a31+

a13a21a32 − a31a22a13−a32a23a11 − a33a21a12

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−

a32a23a11 − a33a21a12

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

a11a22a33 + a12a23a31+a13a21a32 − a31a22a13−a32a23a11 −

a33a21a12

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

∣∣∣∣∣∣∣∣

Now, for n ≥ 4, if let M1j be the corresponding minors to the firstrow, then we have

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

− − − + + +

|A| =n∑

(−1)1+ja1jM1j

|A| =n∑

(−1)i+jaijMij

|A| =n∑

(−1)i+jaijMij

Theorem 7.1

Given an n × n matrix A, if B is an n × n matrix obtained from Aby

1) Adding a multiple of the ith row (column) to the jth row then|B| = |A|

2) Interchanging two consecutive rows (columns), then |B| = −|A|

3) Multiplying a row (column) by a nonzero scalar α then|B| = α|A|

Theorem 7.1

Given an n × n matrix A,

if B is an n × n matrix obtained from Aby

Theorem 7.1

1) Adding a multiple of the ith row (column) to the jth row then

|B| = |A|

Theorem 7.1

2) Interchanging two consecutive rows (columns), then

|B| = −|A|

Theorem 7.1

3) Multiplying a row (column) by a nonzero scalar α then

|B| = α|A|

Theorem 7.1

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

3) If A has a row (column) of zeros, then |A| = 0

4) If A has a two identical rows (columns), then |A| = 0

5) If two rows (columns) of A are proportional, then |A| = 0

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

3) If A has a row (column) of zeros, then

|A| = 0

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

4) If A has a two identical rows (columns), then

|A| = 0

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

5) If two rows (columns) of A are proportional, then

|A| = 0

Theorem 7.2

1) |AT | = |A|

2) |AB| = |A||B|

6) If A is an upper (lower) triangular matrix, then|A| = a11a22a33 · · · ann

Example 7.9

Find the following determinant of the matrix

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

6) If A is an upper (lower) triangular matrix, then

|A| = a11a22a33 · · · ann

Example 7.9

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

Example 7.9

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

Example 7.9

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

Example 7.9

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

Example 7.9

1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ =

(1)(2)(1)(51) = 102

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) =

Solution

∣∣∣∣∣∣∣∣1 −1 2 4−1 3 −2 10 2 1 0−3 1 1 −1

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 2 1 00 −2 7 11

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 7 16

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣1 −1 2 40 2 0 50 0 1 −50 0 0 51

∣∣∣∣∣∣∣∣ = (1)(2)(1)(51) = 102

Theorem 7.3

A matrix A is singular ⇐⇒ |A| = 0 ⇐⇒ Ax = 0 has anonzero solution ⇐⇒ Columns of A are linearly dependent.

Eigenvalues and Eigenvectors.

The equation

Ax = y

can be viewed as a linear transformation that maps (or transforms)a given vector x into a new vector x.

Theorem 7.3

The equation

Ax = y

Theorem 7.3

A matrix A is singular ⇐⇒

|A| = 0 ⇐⇒ Ax = 0 has anonzero solution ⇐⇒ Columns of A are linearly dependent.

The equation

Ax = y

Theorem 7.3

A matrix A is singular ⇐⇒ |A| = 0 ⇐⇒

Ax = 0 has anonzero solution ⇐⇒ Columns of A are linearly dependent.

The equation

Ax = y

Theorem 7.3

A matrix A is singular ⇐⇒ |A| = 0 ⇐⇒ Ax = 0 has anonzero solution ⇐⇒

Columns of A are linearly dependent.

The equation

Ax = y

Theorem 7.3

The equation

Ax = y

Theorem 7.3

The equation

Ax = y

Theorem 7.3

The equation

Ax = y

Theorem 7.3

The equation

Ax = y

Theorem 7.3

The equation

Ax = y

Given an n × n matrix A we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx

but this is equivalent to say that

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Given an n × n matrix A

we consider the problem of finding avector x that is transformed into a multiple of itself

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Given an n × n matrix A we consider the problem of finding avector x

that is transformed into a multiple of itself

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

Ax = λx

Ax = λIx

Ax− λIx = 0

(A− λI) x = 0

The latter equation has nonzero solutions if and only if λ is chosenso that

|A− λI| = 0

This is a polynomial equation of degree n in λ and is called thecharacteristic equation of the matrix A

Values of λ may be either real- or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.

|A− λI| = 0

This is a polynomial equation of degree n in λ

and is called thecharacteristic equation of the matrix A

|A− λI| = 0

This is a polynomial equation of degree n in λ and is called

thecharacteristic equation of the matrix A

|A− λI| = 0

Values of λ may be either real- or complex-valued and are calledeigenvalues of A .

The nonzero vectors that are obtained by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.

|A− λI| = 0

Values of λ may be either real- or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained

by usingsuch a value of λ are called the eigenvectors corresponding tothat eigenvalue.

|A− λI| = 0

Values of λ may be either real- or complex-valued and are calledeigenvalues of A . The nonzero vectors that are obtained by usingsuch a value of λ are called

the eigenvectors corresponding tothat eigenvalue.

|A− λI| = 0

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aandif λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)

are linearly independent.

This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.

a) It is possible to show that if λ1 and λ2

are two eigenvalues of Aandif λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aand

if λ1 6= λ2, then their corresponding eigenvectors x (1) and x (2)

a) It is possible to show that if λ1 and λ2 are two eigenvalues of Aandif λ1 6= λ2, then

their corresponding eigenvectors x (1) and x (2)

This result extends to any set λ1, ..., λk of distinct eigenvalues:

their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.

This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent.

Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.

This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,

then the n eigenvectorsof A , one for each eigenvalue, are linearly independent.

This result extends to any set λ1, ..., λk of distinct eigenvalues:their eigenvectors x (1), ..., x (k) are linearly independent. Thus, ifeach eigenvalue of an n×n matrix is simple,then the n eigenvectorsof A ,

one for each eigenvalue, are linearly independent.

b) On the other hand, if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.

c) In the case of an eigenvalue, λi with multiplicity m, if we canfind m eigenvectors x (i1), x (i2), ..., x (im), linearly independentassociated to λi , we say that the matrix is Non-defective.

d) Otherwise, if we are able to find just x (i1), x (i2), ..., x (iq);q < m linearly independent associated to λi , we say that thematrix is Defective.

b) On the other hand,

if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.

b) On the other hand, if A has one or more repeated eigenvalues,

then there may be fewer than n linearly independent eigenvectorsassociated with A, since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.

b) On the other hand, if A has one or more repeated eigenvalues,then there may be fewer than n linearly independent eigenvectorsassociated with A,

since for a repeated eigenvalue with multiplicitym, we may have q < m linearly independent vectors.

c) In the case of an eigenvalue, λi with multiplicity m, if

we canfind m eigenvectors x (i1), x (i2), ..., x (im), linearly independentassociated to λi , we say that the matrix is Non-defective.

c) In the case of an eigenvalue, λi with multiplicity m, if we canfind m eigenvectors x (i1), x (i2), ..., x (im),

linearly independentassociated to λi , we say that the matrix is Non-defective.

d) Otherwise, if we are able to find

just x (i1), x (i2), ..., x (iq);q < m linearly independent associated to λi , we say that thematrix is Defective.

d) Otherwise, if we are able to find just x (i1), x (i2), ..., x (iq);q < m

linearly independent associated to λi , we say that thematrix is Defective.

Example 7.10

Find the eigenvalues and eigenvectors of the matrix

0 1 11 0 11 1 0

Solution

The eigenvalues λ and eigenvectors x satisfy the equation(A− λI) x = 0, or

Example 7.10

0 1 11 0 11 1 0

Solution

Example 7.10

0 1 11 0 11 1 0

Solution

Example 7.10

0 1 11 0 11 1 0

Solution

Example 7.10

0 1 11 0 11 1 0

Solution

Example 7.10

0 1 11 0 11 1 0

Solution

The eigenvalues λ and

eigenvectors x satisfy the equation(A− λI) x = 0, or

Example 7.10

0 1 11 0 11 1 0

Solution

The eigenvalues λ and eigenvectors x satisfy the equation

(A− λI) x = 0, or

Example 7.10

0 1 11 0 11 1 0

Solution

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

The eigenvalues are the roots of the equation

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ =

− λ3 + 3λ2 + 2 = 0

(A− λI) x =

−λ 1 11 −λ 11 1 −λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣−λ 1 11 −λ 11 1 −λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣1 −λ 1−λ 1 11 1 −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 11 1 −λ−λ 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 −λ 10 λ+ 1 −1− λ0 −λ2 + 1 λ+ 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0−λ λ+ 1 −λ2 + 11 −1− λ λ+ 1

∣∣∣∣∣∣ = − λ3 + 3λ2 + 2 = 0

The roots of are λ1 = 2, λ2 = −1, and λ3 = −1 .

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

We can reduce this to the equivalent system2 −1 −1

0 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

0 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

0 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

0 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

0 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

We can reduce this to the equivalent system2 −1 −10 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

We can reduce this to the equivalent system

2 −1 −10 1 −10 0 0

x1x2x3

1) For λ1 = 2

−λ1 1 11 −λ1 11 1 −λ1

x1x2x3

−2 1 11 −2 11 1 −2

x1x2x3

0 1 −10 0 0

x1x2x3

The above system is reduced immediately to the equations

2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence,

one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable,

let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore

x2 = x3 = α andx1 = (x2 + x3)/2 = α. Thus we have

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

Two equations and three unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore x2 = x3 = α andx1 = (x2 + x3)/2 = α.

Thus we have

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

ααα

; α = real number

2x1 − x2 − x3 = 0 x2 − x3 = 0

ααα

; α = real number

In particular, we have the eigenvector

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

x(1) =

2) For λ2 = −1

−λ2 1 11 −λ2 11 1 −λ2

x1x2x3

1 1 11 1 11 1 1

x1x2x3

The above system is reduced immediately to the single equation

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

−α− β

10−1

01−1

; α, β = real numbers

x1 + x2 + x3 = 0

−α− β

10−1

01−1

x1 + x2 + x3 = 0

−α− β

10−1

01−1

x1 + x2 + x3 = 0

One equation and three unknowns. Hence,

two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

−α− β

10−1

01−1

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables,

let’s say x1 = α, x2 = β, and x3 = −α− β . Thus wehave

−α− β

10−1

01−1

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α,

x2 = β, and x3 = −α− β . Thus wehave

−α− β

10−1

01−1

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and

x3 = −α− β . Thus wehave

−α− β

10−1

01−1

x1 + x2 + x3 = 0

One equation and three unknowns. Hence, two of them are freeveriables, let’s say x1 = α, x2 = β, and x3 = −α− β .

Thus wehave

−α− β

10−1

01−1

x1 + x2 + x3 = 0

−α− β

10−1

01−1

x1 + x2 + x3 = 0

−α− β

10−1

01−1

In this way two linearly independent eigenvectors associated toλ2 = −1 are ( α = β = 1 )

x(2) =

x(3) =

Thus, the three linearly independent eigenvectors, are

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

x(2) =

x(3) =

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ andeigenvectors x satisfy the equation(A− λI) x = 0, or

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

The eigenvalues λ andeigenvectors x satisfy the equation

(A− λI) x = 0, or

Example 7.11

2 −3 −10 −1 0−1 1 2

Solution

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

(A− λI) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣2− λ −3 −1

0 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣ = −

∣∣∣∣∣∣2− λ −3 −1−1 1 2− λ0 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ

2− λ −3 −10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 1 2− λ0 −1− λ (2− λ)2 − 10 −1− λ 0

∣∣∣∣∣∣ =

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

The roots of are λ1 = −1, λ2 = 1, and λ3 = 3 .

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ =

(1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

∣∣∣∣∣∣−1 2− λ 10 (2− λ)2 − 1 −1− λ0 0 −1− λ

∣∣∣∣∣∣ = (1 + λ)[(2− λ)2 − 1

1) For λ1 = −1

(A− λ1I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

3 −3 −10 0 0−1 1 3

x1x2x3

3 −3 −10 0 01 1 3

1 1 30 0 03 −3 −1

1 1 30 0 80 0 0

x1x2x3

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence,

one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable,

let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α.

Therefore we have x1 = −x2 = −α, andx3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have

x1 = −x2 = −α, andx3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, and

x3 = 0 . Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x2 = α. Therefore we have x1 = −x2 = −α, andx3 = 0 .

Thus, we get

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

−αα0

1−10

; α = real number

x1 + x2 + 3x3 = 0 8x3 = 0

−αα0

1−10

; α = real number

A particular eigenvector is

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

x(1) =

1−10

2) For λ2 = 1

(A− λ2I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

x1x2x3

= hspace−2mm

1 −3 −10 −2 0−1 1 1

x1x2x3

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable,

let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α.

Therefore we have x1 = x3 = α, andx2 = 0 . Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have

x1 = x3 = α, andx2 = 0 . Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, and

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

One equation and two unknowns. Hence, one of them is freevariable, let’s say x3 = α. Therefore we have x1 = x3 = α, andx2 = 0 .

Thus, we get

1 −3 −10 −2 00 −2 0

1 −3 −10 −2 00 0 0

x1x2x3

x1 − 3x2 − x3 = 0 − 2x2 = 0

; α = real number

A particular eigenvector is given by

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

; α = real number

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

; α = real number

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

; α = real number

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

; α = real number

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

; α = real number

x(1) =

3) For λ3 = 3

(A− λ3I) x =

2− λ −3 −10 −1− λ 0−1 1 2− λ

=Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

−1 −3 −10 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get

Systems of Linear Algebraic Equations; LinearIndependence, Eigenvalues, Eigenvectors−1 −3 −1

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable,

let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α.

Therefore we have x1 = −x3 = −α, andx2 = 0 . Thus, we get

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have

x1 = −x3 = −α, andx2 = 0 . Thus, we get

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, and

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

One equation and two unknowns. Hence, one of them is a freevariable, let’s say x3 = α. Therefore we have x1 = −x3 = −α, andx2 = 0 .

Thus, we get

0 −4 0−1 1 1

−1 −3 −10 −4 00 4 0

−1 −3 −10 −4 00 0 0

x1x2x3

−x1 − 3x2 − x3 = 0 − 4x2 = 0

−α0α

−101

; α = real number

x(1) =

−101

−α0α

−101

; α = real number

x(1) =

−101

−α0α

−101

; α = real number

x(1) =

−101

−α0α

−101

; α = real number

x(1) =

−101

−α0α

−101

; α = real number

x(1) =

−101

x(1) =

x(2) =

x(3) =

− 101

x(1) =

x(2) =

x(3) =

− 101

x(1) =

x(2) =

x(3) =

− 101

x(1) =

x(2) =

x(3) =

− 101

x(1) =

x(2) =

x(3) =

− 101

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

Example 7.12

4 6 61 3 21 −5 −2

Solution

The eigenvalues λ and eigenvectors x satisfy the equation

(A− λI) x = 0, or

Example 7.12

4 6 61 3 21 −5 −2

Solution

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ =

− λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0

(A− λI) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 2−1 −5 −2− λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣4− λ 6 6

1 3− λ 20 −2− λ −λ

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 3− λ 20 −(4− λ)(3− λ) + 6 6− 2(4− λ)0 −2− λ −λ

∣∣∣∣∣∣ = − λ3 + 5λ2 − 8λ+ 4 =

−(λ− 1)(λ− 2)2 = 0Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

The roots of are λ1 = 1, λ2 = 2, and λ3 = 2 .

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

x1x2x3

3 6 61 2 2−1 −5 −3

x1x2x3

3 6 61 2 20 3 1

1 2 21 2 20 3 1

1 2 21 −3 −10 0 0

x1x2x3

Solving this system yields the eigenvector

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x(1) =

41−3

2) For λ2 = 2

(A− λ2,3I) x =

4− λ 6 61 3− λ 2−1 −5 −2− λ

2 6 61 1 2−1 −5 −4

x1x2x3

2 6 61 1 2−1 −5 −4

1 1 20 4 20 −4 −2

1 1 20 4 20 0 0

x1x2x3

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable, let’s say x3 = α, x2 = 1

2α, and x3 = −2x3 − x2 = −3α .Thus we have

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence,

one of them, is a freevariable, let’s say x3 = α, x2 = 1

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable,

let’s say x3 = α, x2 = 12α, and x3 = −2x3 − x2 = −3α .

Thus we have

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

Two equations and three unknowns. Hence, one of them, is a freevariable, let’s say x3 = α,

x2 = 12α, and x3 = −2x3 − x2 = −3α .

Thus we have

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

2α, and

x3 = −2x3 − x2 = −3α .Thus we have

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

2α, and x3 = −2x3 − x2 = −3α .

Thus we have

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

− 3α12α

− 3α

− 312

x1 + x2 + 2x3 = 0 4x2 + 2x3 = 0

− 3α12α

− 3α

− 312

In this way, there is one linearly independent eigenvector associatedto λ2,3 = 2 , namely,

x(2) =

Therefore, there are just two linearly independent eigenvectors

x(1) =

x(2) =

that is, the matrix A is defective

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

x(1) =

x(2) =

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

Example 7.13

1 0 02 1 −23 2 1

Solution

The eigenvalues λ andeigenvectors x satisfy the equation

(A− λI) x = 0, or

Example 7.13

1 0 02 1 −23 2 1

Solution

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ =

(1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) =

(1− λ)(λ2 − 2λ+ 5) = 0

(A− λI) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

|A− λI| =

∣∣∣∣∣∣1− λ 0 0

2 1− λ −23 2 1− λ

∣∣∣∣∣∣ = (1− λ)

∣∣∣∣1− λ −22 1− λ

∣∣∣∣ =

(1− λ)3 + 4(1− λ) = (1− λ)(λ2 − 2λ+ 5) = 0

The roots of are λ1 = 1, λ2 = 1 + 2 i , and λ3 = 1− 2 i .

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

1) For λ1 = 1

(A− λ1I) x =

1− λ 0 02 1− λ −23 2 1− λ

0 0 02 0 −23 2 0

x1x2x3

2 0 −20 0 03 2 0

1 0 −10 0 03 2 0

1 0 −10 2 40 0 0

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x(1) =

1− 3/2

2) For λ2 = 1 + 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

−2 i 0 02 −2 i −23 2 −2 i

x1x2x3

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have

one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence,

one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α . Hence

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable,

let’s say x2 = α, x3 = −i α . Hence

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α,

x3 = −i α . Hence

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is a free veriable, let’s say x2 = α, x3 = −i α .

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

x1 = 0 − 2 ix2 − 2x3 = 0, 2x2 − 2 ix3 = 0

0α−i α

In this way, there are two real linearly independent eigenvectorassociated to λ2 = 1 + 2 i , namely,

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x(2) =

x(3) =

3) For λ3 = 1− 2 i

(A− λ2I) x =

1− λ 0 02 1− λ −23 2 1− λ

x1x2x3

2 i 0 02 2 i −23 2 2 i

x1x2x3

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is free variable, let’s say x2 = α, x3 = i α . Thus we have

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence,

one ofthem 1 is free variable, let’s say x2 = α, x3 = i α . Thus we have

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is free variable,

let’s say x2 = α, x3 = i α . Thus we have

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is free variable, let’s say x2 = α,

x3 = i α . Thus we have

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

Thus, we have one equation and two unknowns. Hence, one ofthem 1 is free variable, let’s say x2 = α, x3 = i α .

Thus we have

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

x1 = 0; 2i x2 − 2x3 = 0; 2x2 + 2i x3 = 0

0αi α

In this way, there is two real linearly independent eigenvectorassociated to λ2 = 1 + 2 i , namely,

x(2) =

x(3) =

Hence, we have three linearly independent eigenvectors, namely

x(1) =

1− 3/2

x(2) =

x(3) =

that is, the matrix A is Non-defective

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

Hence,

we have three linearly independent eigenvectors, namely

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

x(2) =

x(3) =

x(1) =

1− 3/2

x(2) =

x(3) =

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

Finally, let’s introduce another concept from linear algebra

The Dot Product

For y = (y1, y2, ..., yn), x = (x1, x2, ..., xn) ∈ R, define the dotproduct or inner product or scalar product.as

The Dot Product

Let A be a real-valued n × n matrix.

If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

The Dot Product

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I )

arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

The Dot Product

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A

with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

The Dot Product

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then,

x(R) and x(I ) are two real eigenvectors for Awith eigenvalues λ = u ± i v

The Dot Product

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I )

are two real eigenvectors for Awith eigenvalues λ = u ± i v

The Dot Product

Let A be a real-valued n × n matrix. If x(1) = x(R) ± i x(I ) arecomplex eigenvectors of the matrix A with complex eigenvaluesλ = u ± i v . then, x(R) and x(I ) are two real eigenvectors for A

with eigenvalues λ = u ± i v

The Dot Product

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

a) x and y are said to be orthogonal if < x, y >= 0 .

b) Orthogonal nonzero vectors are linearly independent.

x · y =

< x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=

(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

x · y = < x, y >=(x1 x2 . . . xn

)y1y2...yn

= x1y1 + x2y2 + ...+ xnyn

Theorem 7.10

Let A be an n × n matrix. If A is symetric, ( A = AT ) then

1) All eigenvalues are real.

2) A is always Nondefective.

3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.

Theorem 7.10

3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus

if λ1, λ2, ..., λn are all simple, v1, v2, ..., vn forman orthogonal set.

Theorem 7.10

3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn

are all simple, v1, v2, ..., vn forman orthogonal set.

Theorem 7.10

3) The eigenvectors corresponding to different eigenvalues areorthogonal, thus if λ1, λ2, ..., λn are all simple,

v1, v2, ..., vn forman orthogonal set.

Theorem 7.10

Basic Theory of Systems of First Order LinearEquations

The general theory of a system of n first order linear equations

x ′1 = p11x1 + p12x2 + . . .+ p1nxn + g1(t)x ′2 = p21x1 + p22x2 + . . .+ p2nxn + g2(t)...

...x ′n = pn1x1 + pn2x2 + . . .+ pnnxn + gn(t)

X′ = P(t)X + g(t)

closely parallels that of a single linear equation of nth order.

X′ = P(t)X + g(t)

we assume that P and g are continuous on some intervalα < t < β; that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem 7.4

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.

This is the principle of superposition

we assume that P and g are continuous on some intervalα < t < β;

that is, each of the scalar functionsp11, ..., pnn, g1, ..., gn is continuous there.

Theorem 7.4

If the vector functions x(1) and x(2)

are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.

Theorem 7.4

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 )

then the linear combinationc1x(1) + c2x(2) is also a solution for any constants c1 and c2.

Theorem 7.4

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2)

is also a solution for any constants c1 and c2.

Theorem 7.4

If the vector functions x(1) and x(2) are solutions of thehomogeneus system ( g(t) = 0 ) then the linear combinationc1x(1) + c2x(2) is also a solution

for any constants c1 and c2.

Theorem 7.4

By repeated application of Theorem, we can conclude that ifx(1), ..., x(k) are solutions of the homogeneous system, then

c1x(1) + ...+ ckx(k)

is also a solution for any constants c1, ..., ck .

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

If the vector functions x(1), ..., x(n) are

linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system

for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β,

then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem

can be expressed as a linear combination of x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

If the vector functions x(1), ..., x(n) are linearly independentsolutions of the homogeneous system for each point in the intervalα < t < β, then each solution x = φ(t) of the homogeneoussystem can be expressed as a linear combination of

x(1), ..., x(n) inexactly one way.

φ(t) = c1x(1) + ...+ ckx(k)

c1x(1) + ...+ ckx(k)

Theorem 7.5

φ(t) = c1x(1) + ...+ ckx(k)

If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.

Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < βis saidto be a fundamental set of solutions for that interval.

Theorem 7.6

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by

If the constants c1, ..., cn are thought of as arbitrary,

then theabove equation includes all solutions of the system, and it iscustomary to call it the general solution.

Theorem 7.6

If the constants c1, ..., cn are thought of as arbitrary, then theabove equation

includes all solutions of the system, and it iscustomary to call it the general solution.

Theorem 7.6

If the constants c1, ..., cn are thought of as arbitrary, then theabove equation includes all solutions of the system, and

it iscustomary to call it the general solution.

Theorem 7.6

Any set of solutions x(1), ..., x(n) of the homogeneus system

that islinearly independent at each point in the interval α < t < βis saidto be a fundamental set of solutions for that interval.

Theorem 7.6

Any set of solutions x(1), ..., x(n) of the homogeneus system that islinearly independent at each point in the interval α < t < β

is saidto be a fundamental set of solutions for that interval.

Theorem 7.6

If x(1), ..., x(n)

are solutions of the homogeneus system on theinterval α < t < β, then in this interval W [x(1), ..., x(n)] given by

Theorem 7.6

If x(1), ..., x(n) are solutions of the homogeneus system on theinterval α < t < β,

then in this interval W [x(1), ..., x(n)] given by

Theorem 7.6

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes.To prove this theorem is necessary to establish that

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

∣∣∣∣∣∣∣∣∣∣

either is identically zero or else never vanishes.To prove this theorem is necessary to establish that

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

∣∣∣∣∣∣∣∣∣∣either is identically zero or else never vanishes.

To prove this theorem is necessary to establish that

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

W [x(1), x(2) . . . x(n)] =

∣∣∣∣∣∣∣∣∣∣x(1)1 x

(2)1 . . . x

x(1)2 x

(2)2 . . . x

...... . . .

x(1)n x

(2)n . . . x

dt= [p11 + p22 + ...+ pnn]W

W (t) = ce∫[p11+p22+...+pnn]dt

Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Then, x(1), ..., x(n) form a fundamental set of solutions of thehomogeneous system.

Finally in the case that the solution is complex-valued, we have thefollowing result.

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

Let x(1), ..., x(n)

be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system

thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1),

x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),...,

x(n)(t0) = e(n), respectively, where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

Let x(1), ..., x(n) be the solutions of the homogeneus system thatsatisfy the initial conditions x(1)(t0) = e(1), x(1)(t0) = e(2),..., x(n)(t0) = e(n), respectively,

where t0 is any point inα < t < βand

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · ·

e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

x(1), ..., x(n) form a fundamental set of solutions of thehomogeneous system.

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Then, x(1), ..., x(n)

form a fundamental set of solutions of thehomogeneous system.

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.7

e(1) =

10...0

e(2) =

01...0

· · · e(n) =

00...1

Theorem 7.8

Consider the homogeneous system

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.

Theorem 7.8

X′ = P(t)X

Theorem 7.8

X′ = P(t)X

Theorem 7.8

X′ = P(t)X

Theorem 7.8

X′ = P(t)X

where each element of P is a real-valued continuous function.

Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t) are also solutions of this equation.

Theorem 7.8

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution,

then its real partu(t) and its imaginary part v(t) are also solutions of this equation.

Theorem 7.8

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and

its imaginary part v(t) are also solutions of this equation.

Theorem 7.8

X′ = P(t)X

where each element of P is a real-valued continuous function. Ifx = u(t) + i v(t) is a complex-valued solution, then its real partu(t) and its imaginary part v(t)

are also solutions of this equation.

Theorem 7.8

X′ = P(t)X

Homogeneous Linear Systems with Constant CoefficientsComplex EigenvaluesFundamental MatricesRepeated Eigenvalues

Homogeneous Linear Systems with ConstantCoefficients

We will concentrate most of our attention on systems ofhomogeneous linear equations with constant coefficients

x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.

x′ = Ax

where A is a constant n × n matrix.

Unless stated otherwise, wewill assume further that all the elements of A are real (rather thancomplex) numbers.

x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise,

wewill assume further that all the elements of A are real (rather thancomplex) numbers.

x′ = Ax

where A is a constant n × n matrix. Unless stated otherwise, wewill assume further that all the elements of A

are real (rather thancomplex) numbers.

x′ = Ax

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane.

Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.

x′ = Ax

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax

at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.

x′ = Ax

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and

plotting theresulting vectors, we obtain a direction field of tangent vectors tosolutions of the system of differential equations.

x′ = Ax

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors,

we obtain a direction field of tangent vectors tosolutions of the system of differential equations.

x′ = Ax

The case n = 2 is particularly important and lends itself tovisualization in the x1x2− plane, called the phase plane. Byevaluating Ax at a large number of points and plotting theresulting vectors, we obtain a direction field of tangent vectors to

solutions of the system of differential equations.

x′ = Ax

A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot some solution curves, or trajectories. Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .

A qualitative understanding of the behavior of solutions

can usuallybe gained from a direction field. More precise information resultsfrom including in the plot some solution curves, or trajectories. Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .

A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field.

More precise information resultsfrom including in the plot some solution curves, or trajectories. Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .

A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information results

from including in the plot some solution curves, or trajectories. Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .

A qualitative understanding of the behavior of solutions can usuallybe gained from a direction field. More precise information resultsfrom including in the plot some solution curves, or trajectories.

Aplot that shows a representative sample of trajectories for a givensystem is called a phase portrait .

Now, for the system

x′ = Ax

we look for solutions of the form

x = veλt

where the expon entλ and the vector v are to be determined.Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

where the expon entλ and the vector v are to be determined.

Substituting x in the system gives

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Now, for the system

x′ = Ax

x = veλt

λveλt = Aveλt

(A− λI) v = 0

Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.

If we assume that A is a real-valued matrix, then we must considerthe following possibilities for the eigenvalues of A:

1. All eigenvalues are real and different from each other.

2. Some eigenvalues occur in complex conjugate pairs.

3. Some eigenvalues, either real or complex, are repeated. If the neigenvalues are all real and different, as in the

Thus, to solve the system of differential equations,

we must solvethe above system of algebraic equations. That is, we need to findthe eigenvalues and eigenvectors of the matrix A.

Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations.

That is, we need to findthe eigenvalues and eigenvectors of the matrix A.

Thus, to solve the system of differential equations, we must solvethe above system of algebraic equations. That is, we need to find

the eigenvalues and eigenvectors of the matrix A.

If we assume that A is a real-valued matrix,

then we must considerthe following possibilities for the eigenvalues of A:

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Let’s find the eigenvalues of the matrix A

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

Example 7.14

Consider the system

x′ = Ax =

(1 14 1

Solution

|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒

|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒

|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒

|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒

|A− λI| =

∣∣∣∣1− λ 14 1− λ

∣∣∣∣ = 0

(1− λ)2 − 4 = 0 =⇒

(λ2 − 2λ− 3 = (λ− 3)(λ+ 1) = 0 =⇒

λ1 = 3, λ2 = −1

If λ1 = 3, then the system reduces to the single equation

−2v1 + v2 = 0, =⇒ v2 = 2v1

and a corresponding eigenvector is

v(1) =

λ1 = 3, λ2 = −1

−2v1 + v2 = 0, =⇒ v2 = 2v1

v(1) =

λ1 = 3, λ2 = −1

−2v1 + v2 = 0, =⇒ v2 = 2v1

v(1) =

λ1 = 3, λ2 = −1

−2v1 + v2 = 0, =⇒ v2 = 2v1

v(1) =

λ1 = 3, λ2 = −1

−2v1 + v2 = 0, =⇒ v2 = 2v1

v(1) =

λ1 = 3, λ2 = −1

−2v1 + v2 = 0, =⇒ v2 = 2v1

v(1) =

Similarly, corresponding to λ2 = −1, we find that a correspondingeigenvector is

v(2) =

)The corresponding solutions of the differential equation are

x(1) =

)e3t ; x(2) =

)e−t

v(2) =

x(1) =

)e3t ; x(2) =

)e−t

v(2) =

The corresponding solutions of the differential equation are

x(1) =

)e3t ; x(2) =

)e−t

v(2) =

x(1) =

)e3t ; x(2) =

)e−t

v(2) =

x(1) =

)e3t ;

x(2) =

)e−t

v(2) =

x(1) =

)e3t ; x(2) =

)e−t

The Wronskian of these solutions is

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

Hence the solutions x(1) and x(2) form a fundamental set, and thegeneral solution of the system is

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ =

− 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) =

)e3t + c2

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t +

)e−t

W [x(1), x(2)](t) =

∣∣∣∣ e3t e−t

2e3t −2e−t

∣∣∣∣ = − 4e2t 6= 0

x = c1x(1) + c2x(2) = c1

)e3t + c2

)e−t

Complex Eigenvalues

In this section we consider again a system of n linear homogeneousequations with constant coefficients

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued.

If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt ,

then it follows that λ must be an eigenvalueand v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand

v a corresponding eigenvector of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

where the coefficient matrix A is real-valued. If we seek solutionsof the form x = veλt , then it follows that λ must be an eigenvalueand v a corresponding eigenvector

of the coefficient matrix A.

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex,

we have complex eigenvalues andeigenvectors always appear in complex-conjugate. Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

In the case, λ is complex, we have complex eigenvalues andeigenvectors always appear in complex-conjugate.

Thus, if wehave that

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν;

v±k = a± i b

Complex Eigenvalues

X′ = AX

λ±k = µ± i ν; v±k = a± i b

Complex Eigenvalues

are two complex-conjugate eigenvalues and eigenvectors de lamatrix a, then

X±(t) = e(µ±i ν)t (a± i b)

are complex-valued solutions, but taking in account that

e(µ±i ν)t = eµt (cos(νt)± i sin(νt))

and the principle of superposition, then we have that

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

e(µ±i ν)t =

eµt (cos(νt)± i sin(νt))

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

X2(t) =1

(X+(t)− X−(t)

Complex Eigenvalues

X±(t) = e(µ±i ν)t (a± i b)

X1(t) =1

(X+(t) + X−(t)

)X2(t) =

(X+(t)− X−(t)

Complex Eigenvalues

are two (real) solutions !!!

X1(t) = eµt (acos(νt)− bsin(νt))

X2(t) = eµt (acos(νt) + bsin(νt))

Complex Eigenvalues

Example 7.17

Solve the following ODE

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

Example 7.17

x′ = Ax =

3 1 10 2 10 −1 2

Solution

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣−

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

|A− λI| =

∣∣∣∣∣∣3− λ 1 1

0 2− λ 10 −1 2− λ

∣∣∣∣∣∣ = 0

(3− λ)

∣∣∣∣2− λ 1−1 2− λ

∣∣∣∣− (1)

∣∣∣∣0 10 2− λ

∣∣∣∣+ (1)

∣∣∣∣1 2− λ1 1

∣∣∣∣ =⇒

(3− λ)(λ2 − 4λ+ 6) = 0 =⇒

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2,

λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

λ1 = 2, λ2,3 =4±

√16− (4)(5)

2= 2± i

If λ1 = 3, then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

0 1 10 −1 10 −1 −1

v1v2v3

0 1 10 0 20 0 0

v1v2v3

Complex Eigenvalues

v(1) =

If λ2 = 2 + i , then

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

v(1) =

(A− λ1I) v =

3− λ 1 10 2− λ 10 −1 2− λ

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

3− (2 + i) 1 10 2− (2 + i) 10 −1 2− (2 + i)

v1v2v3

1− i 1 10 −i 10 −1 −i

v1v2v3

1− i 1 10 −i 10 0 0

v1v2v3

1− i 0 1− i0 −i 10 0 0

v1v2v3

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

v(2) =

10−1

x(1) =

e3t ; x(2) = e2t

10−1

cos(t)−

sin(t)

x(3) = e2t

10−1

cos(t) +

sin(t)

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣e3t e2tcos(t) e2tsin(t)0 −e2tsin(t) e2tcos(t)0 −e2tcos(t) −e2tsin(t)

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣1 cos(t) sin(t)0 −sin(t) cos(t)0 −cos(t) −sin(t)

∣∣∣∣∣∣ =

e3te2te2t(sin2(t) + cos2(t)

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

e7t 6= 0

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t

Complex Eigenvalues

W [x(1), x(2), x(3)](t) =

∣∣∣∣∣∣ =

e3te2te2t

∣∣∣∣∣∣ =

)= e7t 6= 0

Complex Eigenvalues

Hence the solutions x(1), x(2) and x(3) form a fundamental set, andthe general solution of the system is

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

X = c1x(1) + c2x(2) + c3x(3) =⇒

X = c1

e3t + c2

10−1

e2tcos(t)−

e2tsin(t)

10−1

e2tcos(t) +

e2tsin(t)

Complex Eigenvalues

x1x2x3

c1e3t + e2t(c2cos(t) + c3sin(t))

0 e2t(−c2sin(t) + c3cos(t))0 −e2t(c2cos(t) + c3sin(t))

Here is the direction field associated with the system

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

x1x2x3

x ′1x ′2x ′3

3 1 10 2 10 −1 2

x1x2x3

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 =

(λ)2 + λ+5

Complex Eigenvalues

Example 7.18

X′ = AX =

(−1/2 1−1 −1/2

Solution

|A− λI| =

∣∣∣∣−1/2− λ 1−1 −1/2− λ

∣∣∣∣ = 0

(−1/2− λ)2 + 1 = (λ)2 + λ+5

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i ,

λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

(−i 1−1 −i

)(v1v2

Complex Eigenvalues

λ1 = −1

2+ i , λ2 = −1

2− i

If λ1 = −12 + i , then

(A− λ1I) x =

(−1/2− λ 1−1 −1/2− λ

)(v1v2

(−1/2− (−1

2 + i) 1−1 −1/2− (−1

2 + i)

)(v1v2

−i 1−1 −i

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

If λ2 = −12 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

v(1) =

)If λ2 = −1

2 − i , then

(A− λ1I) x =

(−1/2− (−1

2 − i) 1−1 −1/2− (−1

2 − i)

)(v1v2

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ;

x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

(i 1−1 i

)(v1v2

v(2) =

(1−i

x(1) =

)e(−1/2+i)t ; x(2) =

)e(−1/2−i)t

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

)Hence a set of real-valued solutions of is

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions,

we can choose the realand imaginary parts of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts

of either x (1) or x (2). In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

To obtain a set of real-valued solutions, we can choose the realand imaginary parts of either x (1) or x (2).

In fact,

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

Hence a set of real-valued solutions of is

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

v(t) = e−t/2(sin(t)cos(t)

Complex Eigenvalues

x(1) =

)e(−1/2+i)t =

)e−t/2 (cos(t) + i sin(t)) =

(e−t/2cos(t)

−e−t/2sin(t)

(e−t/2sin(t)

e−t/2cos(t)

u(t) = e−t/2(

cos(t)−sin(t)

)v(t) = e−t/2

(sin(t)cos(t)

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ e−t/2cos(t) e−t/2sin(t)

−e−t/2sin(t) e−t/2cos(t)

∣∣∣∣ =

e−t/2e−t/2∣∣∣∣ cos(t) sin(t)−sin(t) cos(t)

∣∣∣∣ = e−t 6= 0

Hence the solutions x(1), x(2) form a fundamental set,

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

W [x(1), x(2)](t) =

∣∣∣∣ =

∣∣∣∣ = e−t 6= 0

Complex Eigenvalues

and the general solution of the system is

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system(

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) =

c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

c2e−t/2

(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

Here is the direction field associated with the system(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

)Here is the direction field associated with the system

(x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

X = c1x(1) + c2x(2) = c1e−t/2

(cos(t)−sin(t)

)+ c2e

−t/2(sin(t)cos(t)

x ′1x ′2

(−1/2 1−1 −1/2

)(x1x2

Complex Eigenvalues

Fundamental Matrices

Let’s start with the system

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t)

form a fundamental set ofsolutions on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions

on some interval α < t < β. Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Suppose that x(1)(t), ..., x(n)(t) form a fundamental set ofsolutions on some interval α < t < β.

Then the matrix

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors

x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t),

is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent the matrix is nonsingular.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system.

Since the set ofsolutions is linearly independent the matrix is nonsingular.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

whose columns are the vectors x(1)(t), ..., x(n)(t), is said to be afundamental matrix for the linear system. Since the set ofsolutions is linearly independent

the matrix is nonsingular.

x′ = P(t)x

Ψ(t) =

x(1)1 · · · x

......

x(1)n · · · x

For instance, the system

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

)which are linearly independent. Thus a fundamental matrix for thesystem is

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

which are linearly independent. Thus a fundamental matrix for thesystem is

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

)which are linearly independent.

Thus a fundamental matrix for thesystem is

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

x′ =

(1 14 1

has solutions

x(1)(t) =

); x(2)(t) =

(e−t

−2e−t

Ψ(t) =

(e3t e−t

2e3t −2e−t

Using a fundamental matrix the general solution can be written as

x = Ψ(t)c; c = constant

and if we imposed initial conditions x(t0) = x0, where t0 is a givenpoint on α < t < β and x0 is given initial vector, we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

is the solution of the initial value problem

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

and if we imposed initial conditions

x(t0) = x0, where t0 is a givenpoint on α < t < β and x0 is given initial vector, we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

and if we imposed initial conditions x(t0) = x0,

where t0 is a givenpoint on α < t < β and x0 is given initial vector, we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

and if we imposed initial conditions x(t0) = x0, where t0 is a givenpoint on α < t < β and

x0 is given initial vector, we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

and if we imposed initial conditions x(t0) = x0, where t0 is a givenpoint on α < t < β and x0 is given initial vector,

we obtain

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x;

x(t0) = x0

Ψ(t0)c = x0

c = Ψ−1(t0)x0

x = Ψ(t)Ψ−1(t0)x0

x′ = P(t)x; x(t0) = x0Dr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

Sometimes it is convenient to make use of the specialfundamental matrix , denoted by Φ, such that the initialcondition

x(j) = e(j); e(j) =

0...1...0

j − th row

Recall that each column

of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Recall that each column of the fundamental matrix Ψ(t)

is asolution of the ODE. It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE.

It follows that Ψ(t) satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Recall that each column of the fundamental matrix Ψ(t) is asolution of the ODE. It follows that Ψ(t)

satisfies the matrixdifferential equation

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

Sometimes it is convenient

to make use of the specialfundamental matrix , denoted by Φ, such that the initialcondition

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

Sometimes it is convenient to make use of the specialfundamental matrix ,

denoted by Φ, such that the initialcondition

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j);

e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Ψ′ = P(t)Ψ

x(j) = e(j); e(j) =

0...1...0

j − th row

Thus Φ(t) has the property that

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

and the solution of the IVP is given by

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 =

Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) =

Ψ(t)Ψ−1(t0)

Φ(t0) =

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

x = Φ(t)Φ−1(t0)x0 = Φ(t)x0

in another words

Φ(t) = Ψ(t)Ψ−1(t0)

Thus, for instance the system

x′ =

(1 14 1

subject to the different initial conditions

x(1)(0) =

); x(2)(0) =

)has the particular solutions equal to

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

has the particular solutions equal to

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x′ =

(1 14 1

x(1)(0) =

); x(2)(0) =

x(t) =1

(e−t

−2e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

Note that the elements of Φ(t) are more complicated than those ofthe fundamental matrix Ψ(t); however, it is now easy to determinethe solution corresponding to any set of initial conditions.

x(t) =

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

Note that the elements of Φ(t)

are more complicated than those ofthe fundamental matrix Ψ(t); however, it is now easy to determinethe solution corresponding to any set of initial conditions.

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

Note that the elements of Φ(t) are more complicated

than those ofthe fundamental matrix Ψ(t); however, it is now easy to determinethe solution corresponding to any set of initial conditions.

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

Note that the elements of Φ(t) are more complicated than those ofthe fundamental matrix Ψ(t);

however, it is now easy to determinethe solution corresponding to any set of initial conditions.

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

Note that the elements of Φ(t) are more complicated than those ofthe fundamental matrix Ψ(t); however, it is now easy to determinethe solution

corresponding to any set of initial conditions.

x(t) =1

)− 1

(e−t

−2e−t

)Hence

Φ(t) =

3t + 12e−t 1

2e3t − 1

2e−t

e3t − e−t 12e

3t + 12e−t

The Matrix eAt

Recall that the solution of the initial value problem

x ′ = ax , x(0) = x0, a = constant

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem for an n × nsystem

x′ = Ax, x(0) = x0

where A is a constant matrix.

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

Now, consider the corresponding initial value problem

for an n × nsystem

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

The Matrix eAt

is given by

x(t) = x0eat

x′ = Ax, x(0) = x0

Applying the results of already obtained, we can write its solutionas

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

The scalar exponential function eat can be represented by thepower series

eat = 1 +∞∑n=1

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series

Applying the results of already obtained,

we can write its solutionas

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

where Φ(0) = I.

Thus, Φ(t), is playing the roll of eat . let’s seethis with more detail.

eat = 1 +∞∑n=1

x = Φ(t)x0

where Φ(0) = I. Thus, Φ(t), is playing the roll of eat .

let’s seethis with more detail.

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

The scalar exponential function eat

can be represented by thepower series

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

x = Φ(t)x0

eat = 1 +∞∑n=1

which converges for all t.

Let us now replace the scalar a by then × n constant matrix A and consider the corresponding series

x = Φ(t)x0

eat = 1 +∞∑n=1

which converges for all t. Let us now replace the scalar a by then × n constant matrix A and

consider the corresponding series

x = Φ(t)x0

eat = 1 +∞∑n=1

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

By differentiating the above series term by term, we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

I + At +A2t2

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I +

At +A2t2

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

...+Ant2

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix.

It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat

each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges

for all t asn→∞. Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞.

Thus, we have a well defined n × n matrix, which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

Each term in the series is an n × n matrix. It is possible to showthat each element of this matrix sum converges for all t asn→∞. Thus, we have a well defined n × n matrix,

which will bedenote by eAt

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

By differentiating the above series

term by term, we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

By differentiating the above series term by term,

we obtain

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!=

∞∑n=1

]= AeAt

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

I +∞∑n=1

n!= I + At +

2!+ ...+

n!+ ...

eAt = I +∞∑n=1

=∞∑n=1

Antn−1

(n − 1)!= A

∞∑n=1

]= AeAt

Therefore, eAt satisfies the differential equation

= AeAt

Further, by setting t = 0 in the definition of eAt we find that eAt

satisfies the initial condition

eAt∣∣∣t=0

In this way, we have that the fundamental matrix Φ satisfies thesame initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

Further, by setting t = 0

in the definition of eAt we find that eAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

Further, by setting t = 0 in the definition of eAt

we find that eAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

In this way,

we have that the fundamental matrix Φ satisfies thesame initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

In this way, we have that

the fundamental matrix Φ satisfies thesame initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

In this way, we have that the fundamental matrix Φ

satisfies thesame initial value problem as eAt , namely,

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

In this way, we have that the fundamental matrix Φ satisfies thesame initial value problem as

eAt , namely,

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

= AeAt

eAt∣∣∣t=0

Φ′ = AΦ, Φ(0) = I

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the fundamental matrix Φ(t)are the same. Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

Then, by uniqueness of an IVP

(extended to matrix differentialequations), we conclude that eAt and the fundamental matrix Φ(t)are the same. Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

Then, by uniqueness of an IVP (extended to matrix differentialequations),

we conclude that eAt and the fundamental matrix Φ(t)are the same. Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and

the fundamental matrix Φ(t)are the same. Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the fundamental matrix Φ(t)are the same.

Thus we can write the solution of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

Then, by uniqueness of an IVP (extended to matrix differentialequations), we conclude that eAt and the fundamental matrix Φ(t)are the same. Thus we can write the solution

of the initial valueproblem

x = Ax, x(0) = x0

in the form

x = eAtx0

x = Ax, x(0) = x0

in the form

x = eAtx0

x = Ax, x(0) = x0

in the form

x = eAtx0

x = Ax, x(0) = x0

in the form

x = eAtx0

x = Ax, x(0) = x0

in the form

x = eAtx0

Diagonalizable Matrices.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty is that the equations are usuallycoupled.

Hence the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

The basic reason why a system of linear (algebraic or differential)equations presents some difficulty

is that the equations are usuallycoupled.

Hence the equations in the system must be solved simultaneously.

On the contrary, if the system is uncoupled, then each equationcan be solved independently of all the others.

Hence the equations in the system must be solved simultaneously.On the contrary, if the system is uncoupled,

then each equationcan be solved independently of all the others.

Transforming the coupled system into

an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable )

corresponds to transforming the coefficient matrix A intoa diagonal matrix. Eigenvectors are useful in accomplishing such atransformation.

Transforming the coupled system into an equivalent uncoupledsystem ( in which each equation contains only one unknownvariable ) corresponds to transforming the coefficient matrix A intoa diagonal matrix.

Eigenvectors are useful in accomplishing such atransformation.

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

Ax(1) = λ1x(1); Ax(2) = λ2x(2); ...Ax(n) = λnx(n)

and considering the matrix

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Let’s assume that the matrix A

has n eigenvectors x(1), x(2), ...,x(n) linearly indepedent, then

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

Let’s assume that the matrix A has n eigenvectors x(1), x(2), ...,x(n)

linearly indepedent, then

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

x(1)1 · · · x(n)

......

x(1)n · · · x

we have

Ax(1) · · · Ax

......

... · · ·...

(1)1 · · · λnx

λ1x(1)2

λ1x(1)n λnx

where D is the diagonal matrix

λ2. . .

whose diagonal elements are the eigenvalues of A. From the lastequations we have that it follows that

T−1AT = D

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix by the process shown in theabove equation.

λ2. . .

T−1AT = D

λ2. . .

T−1AT = D

λ2. . .

T−1AT = D

λ2. . .

T−1AT = D

λ2. . .

T−1AT = D

λ2. . .

T−1AT = D

Thus, if the eigenvalues and eigenvectors of A are known,

A canbe transformed into a diagonal matrix by the process shown in theabove equation.

λ2. . .

T−1AT = D

Thus, if the eigenvalues and eigenvectors of A are known, A canbe transformed into a diagonal matrix

by the process shown in theabove equation.

λ2. . .

T−1AT = D

This process is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix T such that T−1AT = D. Inthis case, A is not diagonalizable.

Fundamental MatricesThis process is known as a similarity transformation.

Alternatively, we may say that A is diagonalizable.

Fundamental MatricesThis process is known as a similarity transformation.Alternatively, we may say that A is diagonalizable.

If A is Hermitian,

then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.

The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n)

of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal,

so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that

they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1

for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i .

It is easy verify that T−1 = T∗. Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗.

Inother words, the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words,

the inverse of T is the same as its adjoint (thetranspose of its complex conjugate).

If A is Hermitian, then the determination of T−1 is very simple.The eigenvectors v(1), ..., v(n) of A are known to be mutuallyorthogonal, so let us choose them so that they are also normalizedby < v(i), v(i) >= 1 for each i . It is easy verify that T−1 = T∗. Inother words, the inverse of T

is the same as its adjoint (thetranspose of its complex conjugate).

Finally,

we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix T such that T−1AT = D. Inthis case, A is not diagonalizable.

Finally, we note that if A has fewer than n linearly independenteigenvectors,

then there is no matrix T such that T−1AT = D. Inthis case, A is not diagonalizable.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix T

such that T−1AT = D. Inthis case, A is not diagonalizable.

Finally, we note that if A has fewer than n linearly independenteigenvectors, then there is no matrix T such that T−1AT = D. Inthis case,

A is not diagonalizable.

Repeated Eigenvalues

We conclude our consideration of the linear homogeneous systemwith constant coefficients

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

Then λ is an eigenvalue of algebraic multiplicity 2 of the matrix A.In this event, there are two possibilities: The matrx A isnon-defectine and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues.

suppose that λ is a repetead root of the characteristicequation ∣∣∣A− λI

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

with a discussion of the case in which the matrix A has a repeatedeigenvalues. suppose that λ is a repetead root of the characteristicequation

∣∣∣A− λI∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then λ is an eigenvalue of algebraic multiplicity 2 of the matrix A.

In this event, there are two possibilities: The matrx A isnon-defectine and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then λ is an eigenvalue of algebraic multiplicity 2 of the matrix A.In this event, there are two possibilities:

The matrx A isnon-defectine and there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

Then λ is an eigenvalue of algebraic multiplicity 2 of the matrix A.In this event, there are two possibilities: The matrx A isnon-defectine and

there is still a fundamental set of solutions ofthe form

{veλt ,weλt

x′ = Ax

∣∣∣ = 0

{veλt ,weλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation has a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

In this way, it may be natural to attempt to find a secondindependent solution of the form

x = wteλt

However,

if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective,

there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution

ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt

associated with this eigenvalue. Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue.

Therefore, toconstruct the general solution, it is necessary to find other solutionof a different form.

x = wteλt

However, if the matrx A is defective, there is just one solution ofthe form veλt associated with this eigenvalue. Therefore, toconstruct the general solution,

it is necessary to find other solutionof a different form.

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0

when the characteristic equation has a doubleroot r . In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation has a doubleroot r .

In that case we found one exponential solution y1(t) = ert ,but a second independent solution had the form y2(t) = tert

x = wteλt

Recall that a similar situation occurred for the linear equationay ′′ + by ′ + cy = 0 when the characteristic equation has a doubleroot r . In that case we found one exponential solution y1(t) = ert ,

but a second independent solution had the form y2(t) = tert

x = wteλt

In this way,

it may be natural to attempt to find a secondindependent solution of the form

x = wteλt

In this way, it may be natural to attempt to find a secondindependent solution

of the form

x = wteλt

but, doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

and substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and only thesecond one is remaining to be solved.

doing this and substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and

substituting x in the system we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and substituting x in the system

we find thatw = 0. Thus, we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

but, doing this and substituting x in the system we find thatw = 0. Thus,

we propose

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

substituting this new x in the system we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

and substituting this new x in the system

we find the system

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

The first equation is already solved with w = v and

only thesecond one is remaining to be solved.

x = wteλt + ueλt

(A− λI) w = 0

(A− λI) u = w

Example 7.19

Find the solution of the system

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

Example 7.19

x′ = Ax =

(1 −11 3

Solution

|A− λI| =

∣∣∣∣1− λ −11 3− λ

∣∣∣∣ = 0

(λ− 1)(λ− 3) + 1 = 0 =⇒ (λ− 2)2 = 0

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

)and a corresponding eigenvector is

v(1) =

λ1 = 2,

λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

λ1 = 2, λ2 = 2,

If λ1,2 = 2, then

(A− λ1,2I) v =

(1− λ −1

1 3− λ

)(v1v2

(−1 −11 1

)(v1v2

v(1) =

and the solution is

x(1) =

Now, for the second solution we propose

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t +

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u =

(A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u =

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

and the solution is

x(1) =

x(2) = vte2t + ue2t

where u satisfies

(A− λI) u = (A− 2I) u = v

(−1 −11 1

)(u1u2

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

)then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

so if u1 = k ,

where k is arbitrary, then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary,

then u2 = −k − 1. If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

so if u1 = k , where k is arbitrary, then u2 = −k − 1.

If we write

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

then by substituting for w and u, we obtain

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t +

(1−1

we have

−u1 − u2 = 1

−1− k

(0−1

(1−1

x(2) =

(1−1

)te2t +

(0−1

)e2t + k

(1−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)

}form a fundamental set of solutions of

the system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above

is merely a multiple of the first solutionx (1)(t) and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t)

and may be ignored, but the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

The last term above is merely a multiple of the first solutionx (1)(t) and may be ignored, but

the first two terms constitute anew solution:

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation

shows that W [x (1), x (2)](t) = − e4t 6= 0and therefore

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) =

− e4t 6= 0and therefore

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

An elementary calculation shows that W [x (1), x (2)](t) = − e4t 6= 0

and therefore{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

form a fundamental set of solutions ofthe system. The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

the system.

The general solution is

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) +

c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

x(2) =

(1−1

)te2t +

(0−1

{x (1), x (2)

x = c1x(1) + c2x(2) = c1

(1−1

)e2t + c2

((1−1

)te2t +

(0−1

Consider again the system

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but that thereis only one corresponding eigenvector v. Then one solution is

x(1)(t) = veλtwhere v satisfies

(A− λI) v = 0

and a second solution is given by

x(2)(t) = vteλt + ueλt

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A,

but that thereis only one corresponding eigenvector v. Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

and suppose that r = λ is a double eigenvalue of A, but that thereis only one corresponding eigenvector v.

Then one solution is

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

x(1)(t) = veλt

where v satisfies

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = v

x′ = Ax

(A− λI) v = 0

where u satisfies

(A− λI) u = vDr. Marco A Roque Sol Ordinary Differential Equations. Session 7

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

The vector u is known as a generalized eigenvector.

Even though |A− λI| = 0,

it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u

( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .

Now, Using the above equation, together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation,

together with the equation for v,we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Even though |A− λI| = 0, it can be shown that it is alwayspossible to solve it for u ( Actually, there are infinetly solutions ) .Now, Using the above equation, together with the equation for v,

we get

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI)

[(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u =

(A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u =

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

(A− λI) [(A− λI) u = v]

(A− λI)2 u = (A− λI) v

(A− λI)2 u = 0

Fundamental matrices are formed by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix forthe example 7.19 can be formed from the solutions x (1)(t) andx (2)(t) obtained before :

Ψ(t) =

(e2t te2t

−e2t −te2t − e2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation Φ(t) = Ψ(t)Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed

by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix forthe example 7.19 can be formed from the solutions x (1)(t) andx (2)(t) obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed by arranging linearly independentsolutions in columns.

Thus, for example, a fundamental matrix forthe example 7.19 can be formed from the solutions x (1)(t) andx (2)(t) obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix

forthe example 7.19 can be formed from the solutions x (1)(t) andx (2)(t) obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix forthe example 7.19

can be formed from the solutions x (1)(t) andx (2)(t) obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix forthe example 7.19 can be formed from the solutions x (1)(t) and

x (2)(t) obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Fundamental matrices are formed by arranging linearly independentsolutions in columns. Thus, for example, a fundamental matrix forthe example 7.19 can be formed from the solutions x (1)(t) andx (2)(t)

obtained before :

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation Φ(t) = Ψ(t)Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = I

can also be found from the relation Φ(t) = Ψ(t)Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation

Φ(t) = Ψ(t)Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation Φ(t) =

Ψ(t)Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation Φ(t) = Ψ(t)

Ψ−1(0). Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

)In particular, the fundamental matrix Φ(t) that satisfies Φ(0) = Ican also be found from the relation Φ(t) = Ψ(t)Ψ−1(0).

Thus, inthis case

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Ψ(t) =

(e2t te2t

)= e2t

(1 t−1 −1− t

Ψ(0) =

(1 0−1 −1

)=⇒ Ψ−1(0) =

(1 0−1 −1

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as the exponential matrix eAt .

Jordan Canonical Forms

An n × n matrix A can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Φ(t) =

Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) =

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

(1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) =

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

The latter matrix is also known as the exponential matrix eAt .

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix

is also known as the exponential matrix eAt .

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

)The latter matrix is also known as

the exponential matrix eAt .

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

An n × n matrix A

can be diagonalized only if it has a fullcomplement of n linearly independent eigenvectors.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

An n × n matrix A can be diagonalized only if

it has a fullcomplement of n linearly independent eigenvectors.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

An n × n matrix A can be diagonalized only if it has a fullcomplement of

n linearly independent eigenvectors.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

If there is a shortage of eigenvectors

(because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

If there is a shortage of eigenvectors (because of repeatedeigenvalues),

then A can always be transformed into a nearlydiagonal matrix called its Jordan form.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed

into a nearlydiagonal matrix called its Jordan form.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

If there is a shortage of eigenvectors (because of repeatedeigenvalues), then A can always be transformed into a nearlydiagonal matrix called

its Jordan form.

Φ(t) = Ψ(t)Ψ−1(0) = e2t(

1 t−1 −1− t

) (1 0−1 −1

Φ(t) = e2t(

1− t −tt 1 + t

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

A Jordan form, J,

has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

A Jordan form, J, has the eigenvalues of A

on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

A Jordan form, J, has the eigenvalues of A on the main diagonal,

ones in certain positions on the diagonal above the main diagonal,and zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

A Jordan form, J, has the eigenvalues of A on the main diagonal,ones in certain positions on the diagonal above the main diagonal,and

zeros elsewhere.

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

J(t) =

λ1 10 λ1 10 0 λ1

λ2 10 λ2

λ3. . .

λn 10 λn

Consider again the matrix A given by the equation

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form,

we construct thetransformation matrix T with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T

with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T with the single eigenvector v

in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T with the single eigenvector v in its firstcolumn and

the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column.

Then T and its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

To transform A into its Jordan form, we construct thetransformation matrix T with the single eigenvector v in its firstcolumn and the generalized eigenvector u ( k = 0 ) in thesecond column. Then T and

its inverse are given by

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

(1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

(1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

x′ = Ax =

(1 −11 3

(1 0−1 −1

)T−1 =

(1 0−1 −1

)It follows that

J = T−1AT =

(1 0−1 −1

) (1 −11 3

) (1 0−1 −1

(2 10 2

Finally, If we start again from

x′ = Ax =

(1 −11 3

the transformation x = Ty where T is given above, produces thesystem

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

the transformation x = Ty

where T is given above, produces thesystem

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

the transformation x = Ty where T is given above,

produces thesystem

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2,

y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t ,

y1 = c1te2t + c2e

x′ = Ax =

(1 −11 3

J′ = Jy

y ′1 = 2y1 + y2, y ′2 = 2y2

y2 = c1e2t , y1 = c1te

2t + c2e2t

Thus, two independent solutions of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

and the corresponding fundamental matrix is

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

−e2t −e2t − te2t

)which is the same as the fundamental matrix that we obtainedbefore.

two independent solutions of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

Thus, two independent solutions

of the y−system are

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ;

y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I,

we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt .

To obtaina fundamental matrix for the original system, we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix

for the original system, we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

)Since Ψ̂(0) = I, we can also identify this matrix as eJt . To obtaina fundamental matrix for the original system,

we now form theproduct

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) =

TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

which is the same as the fundamental matrix that we obtainedbefore.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

)which is the same

as the fundamental matrix that we obtainedbefore.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

)which is the same as the fundamental matrix

that we obtainedbefore.

y(1)(t) =

)e2t ; y(2)(t) =

Ψ̂(t) =

(e2t te2t

Ψ(t) = TeJt =

(e2t te2t

Ordinary Di erential Equations. Session 7roquesol/Math_308_Fall_2017_Session_7.pdfIntroduction...

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