Instructor : Ferry Wahyu Wibowo, S.Si., M.Cs.

Linear Algebra and Matrices

Meeting : V

V.1 The Inverse of a Matrix

DefinitionLet A be an n n matrix. If a matrix B can be found such that AB = BA = In,

then A is said to be invertible and B is called the inverse of A. If such a matrix

B does not exist, then A has no inverse. (denote B = A 1, and A k=(A 1)k )

4321

A

Example 1

Prove that the matrix has inverse .12

2

1

2

3B

Proof

22

1

2

3 100112

4321

IAB

22

1

2

3 1001

432112

IBA

Thus AB = BA = I2, proving that the matrix A has inverse B.

Theorem V.1

If a matrix has an inverse, that inverse is unique.

Proof

Let B and C be inverses of A.

Thus AB = BA = In, and AC = CA = In.

Multiply both sides of the equation AB = In by C.

C(AB) = CIn

(CA)B = C

InB = C

B = C

Thus an invertible matrix has only one inverse.

Let A be an invertible n n matrix

How to find A-1?

. ,Let 2121

1

nnn CCCIXXXA

We shall find A 1 by finding X1, X2, …, Xn.

Since AA 1 =In, then .2121 nn CCCXXXA

.,, , i.e., 2211 nn CAXCAXCAX

Solve these systems by using Gauss-Jordan elimination:

.:

::matrix augmented

21

21

nn

n

XXXI

CCCA

.:: 1AIIA nn

Gauss-Jordan Elimination for

finding the Inverse of a MatrixLet A be an n n matrix.

1. Adjoin the identity n n matrix In to A to form the matrix [A : In].

2. Compute the reduced echelon form of [A : In].

If the reduced echelon form is of the type [In : B], then B is the inverse of A.

If the reduced echelon form is not of the type [In : B], in that the first n n

submatrix is not In, then A has no inverse.

An n n matrix A is invertible if and only if its reduced echelon form is In.

Example 1

Determine the inverse of the matrix

531532211

A

Solution

100531010532001211

]:[ 3IA101320012110001211

1R

R1R3

2)(R2

101320012110001211

R2)1(

123100012110013101

R2)2(R3R2R1

.123135110

Thus, 1A

123100135010010001

R3)1(R2R3R1

1

135000011210012301

3R2R3

R2)1(R1

Example 2Determine the inverse of the following matrix, if it exist.

412721511

A

Solution

100412010721001511

]:[ 3IA102630011210001511

R1)2(R3

1)R1(R2

There is no need to proceed further.

The reduced echelon form cannot have a one in the (3, 3) location.

The reduced echelon form cannot be of the form [In : B].

Thus A–1 does not exist.

Properties of Matrix Inverse

Let A and B be invertible matrices and c a nonzero scalar, Then

AA 11)( 1.11 1

)( .2 Ac

cA

111)( .3 ABAB

nn AA )()( .4 11

tt AA )()( .5 11

Proof

1. By definition, AA 1=A 1A=I.

))(())(( .2 11 11cAAIAcA

cc

))(( )())(( .3 1111111 ABABIAAABBAABAB

nn

nn

nn AAIAAAAAA )( )( .4 1

times

11

times

1

,)( )( ,

,)( )( , .5

111

111

IAAAAIAA

IAAAAIAA

ttt

ttt

Example 4

.)( compute n toinformatio

this Use.4311

shown that becan it then ,1314

If

1

1

tA

AA

Solution

.)()(41

31

43

1111

t

tt AA

Theorem V.2

Let AX = Y be a system of n linear equations in n variables.

If A–1 exists, the solution is unique and is given by X = A–1Y.

Proof

(X = A–1Y is a solution.)

Substitute X = A–1Y into the matrix equation.

AX = A(A–1Y) = (AA–1)Y = InY = Y.

(The solution is unique.)

Let X1 be any solution, thus AX1 = Y. Multiplying both sides of this equation by

A–1 gives

A–1AX1= A–1Y

InX1 = A–1Y

X1 = A–1Y.

Example 3

Solve the system of equations

253

3532

12

321

321

321

xxx

xxx

xxx

Solution

This system can be written in the following matrix form:

231

531532211

3

2

1

x

x

x

If the matrix of coefficients is invertible, the unique solution is

231

531532211

1

3

2

1

x

x

x

This inverse has already been found in Example 2. We get

121

231

123135110

3

2

1

x

x

x

.1 ,2 ,1 issolution unique The 321 xxx

Elementary Matrices

DefinitionAn elementary matrix is one that can be obtained from the identity matrix

In through a single elementary row operation.

Example

100

010

001

3I

010

100

001

1ER2 R3

100

050

001

2E5R2

100

012

001

3ER2+ 2R1

Elementary Matrices

ihg

fed

cba

A

AEA

fed

ihg

cba

1

010

100

001

R2 R3

AEA

ihg

fed

cba

2

100

050

001

5555R2

AEA

ihg

cfbead

cba

3

100

012

001

222R2+ 2R1

Notes for elementary matrices

Each elementary matrix is square and invertible.

Example

If A and B are row equivalent matrices and A

is invertible, then B is invertible.

Proof

If A … B, then

B=En … E2 E1 A for some elementary matrices En, … , E2 and E1.

So B 1 = (En … E2 E1A) 1 =A 1E11 E2

1 … En1.

1221EI

RR

100

010

021

1E

100

010

001

I

100

010

021

2E

, 221

1 IERR

IEE 12 i.e.,

Homework

dc

baA

Exercise 1

If , show that .)(

11

ac

bd

bcadA

Exercise 2

Prove that (AtBt) 1 = (A 1B 1)t.

Exercise 3

True or False:

(a) If A is invertible A 1 is invertible.

(b) If A is invertible A2 is invertible.

(c) If A has a zero on the main diagonal it is not invertible.

(d) If A is not invertible then AB is not invertible.

(e) A 1 is row equivalent to In.

V.2 Matrix Transformations,

Rotations, and DilationsA function, or transformation, is a rule that assigns to each element of a set a

unique element of another set.

We will be especially interested in linear transformations, which are

transformations that preserve the mathematical structure of a vector space.

Consider the function f(x) =3x2+4.

domain of the function: the set of all possible x

f(2)=16, we say that the image of 2 is 16.

Extend these ideas to functions between vector spaces

We usually use the term transformation rather than function in linear algebra.

Consider the transformation T maps R3 into R2 defined by

T(x, y, z) = (2x, y z)

The domain of T is R3 and we say the codomain is R2.

T(1, 4, 2) = (2, 6) The image of (1, 4, 2) is (2, 6).

Convenient representations:

zyx

z

yx

T2 , and the image of .

62

is 2

41

DefinitionA transformation T of Rn into Rm is a rule that assigns to each vector u in Rn a

unique vector v in Rm. Rn is called the domain of T and Rm is the codomain.

We write T(u) = v; v is the image of u under T. The term mapping is also used

for a transformation.

Dilationand Contraction

Consider the transformation , where r > 0. yx

ryx

T

Figure V.1

This equation can be written as the following useful matrix form.

yx

rr

yx

T0

0

Reflection

Consider the transformation . T is called a reflection.y

xyx

T

This equation can be written as the following useful matrix form.

yx

yx

T10

01

Figure V.2

Matrix transformationsEvery matrix defines a transformation. Let A be a matrix and x be a column

vector such that Ax exists. Then A defines the matrix transformation T(x) = Ax.

For example, defines 140235

Az

yx

z

yx

T140235

86

431

T2

14

213

Tand

Figure V.3 Matrix transformations.

214

213, written as

86

431, written as

DefinitionLet A be an m n matrix. Let x be an element of Rn written in a column matrix

form. A defines a matrix transformation T(x)=Ax of Rn into Rm. The vector Ax is

the image of x. The domain of the transformation is Rn and codomain is Rm.

We write T: Rn Rm if T maps Rn into Rm.

Geometrical Properties:

Matrix transformations maps line segments into line segments (or points). If the

matrix is invertible the transformation also maps parallel lines into parallel lines.

Example 1Consider the transformation T: R2 R2 defined by the matrix .

Determine the image of the unit square under this transformation.

32

24A

Sol.

0

0 ,

1

0 ,

1

1 ,

0

1ORQP

The unit square is the square whose vertices are the points

3

2

1

0

32

24 ,

5

6

1

1

32

24 ,

2

4

0

1

32

24

Since

0

0

0

0 ,

3

2

1

0 ,

5

6

1

1 ,

2

4

0

1

OOR'RQ'QP'P

The images are:

Figure V.4

The square PQRO is mapped into the parallelogram P’Q’R’O.

Composition of Transformations

Figure V.5

Consider the matrix transformations T1(x)=A1x and T2(x)=A2x.

The composite transformation T=T2 T1 is given by

T(x) = T2(T1(x)) = T2 (A1x) =A2A1x

Thus T is defined by the matrix product A2A1.

T(x) = A2A1x

V.3 Linear TransformationsA vector space has two operations: addition and scalar multiplication.

Consider the matrix transformation T(u)=Au.

T(u+v) = A(u+v) = Au+Av = T(u)+T(v)

DefinitionLet u and v be vectors in Rn and let c be a scalar. A transformation

T: Rn → Rm is said to be a linear transformation if

T(u + v) = T(u) + T(v) and T(cu) = cT(u).

uT T(u)

T(v)

v

u+v T(u+v) = T(u)+T(v)

ucu T

T(cu) = cT(u)

T(u)

T(cu) = A(cu) = cAu = cT(u)

Example 1Prove that the following transformation T: R2 → R2 is linear.

T(x, y) = (x y, 3x)

Solution

“+”: Let (x1, y1) and (x2, y2) R2. Then

T((x1, y1) + (x2, y2)) = T(x1 + x2, y1 + y2)

= (x1 + x2 y1 y2, 3x1 + 3x2)

= (x1 y1, 3x1) + (x2 y2, 3x2)

= T(x1, y1) + T(x2, y2)

“ ”: Let c be a scalar.

T(c(x1, y1)) = T(cx1, cy1) = (cx1 cy1, 3cx1)

= c(x1 y1, 3x1)

= cT(x1, y1)

Thus T is linear.

Note A linear transformation T: U → U is called a linear operator.

Example 2Show that the following transformation T: R3 → R2 is not linear.

T(x, y, z) = (xy, z)

Solution

“+”: Let (x1, y1, z1) and (x2, y2, z2) R3. Then

T((x1, y1, z1) + (x2, y2, z2))

= T(x1 + x2 , y1 + y2 , z1 + z2)

= ((x1 + x2)(y1 y2), z1 + z2)

T is not linear.

and T(x1, y1, z1) + T(x2, y2, z2) = (x1y1, z1) + (x2y2, z2)

Thus, in general

T((x1, y1, z1) + (x2, y2, z2)) T(x1, y1, z1) + T(x2, y2, z2).

Example 3Determine a matrix A that describes the linear transformation

y

yxyx

T3

2)(

Solution

It can be shown that T is linear. The domain of T is R2.

We find the effect of T on the standard basis of R2.

02

)01

(T and31

)10

(T

3012

A

T can be written as

yx

yx

T3012

)(

Why? See the next page.

Let T: Rn → Rm be a linear transformation. {e1, e2, …, en} be the standard basis

of Rn, and u be an arbitrary vector in Rn.

n

n

a

a

1

1

1

0

,,

0

1

uee

Matrix Representation

Express u in terms of the basis, u = a1e1+a2e2+ … + anen.Since T is a linear transformation,

T(u) = T(a1e1+ a2e2 + … + anen) = T(a1e1) + T(a2e2) + …+ T(anen)

= a1T(e1) + a2T(e2) + …+ anT(en)

])()()([

1

21

n

n

a

a

TTT eee

Thus the linear transformation T is defined by the matrix

A = [ T(e1) … T(en) ].

A is called the standard matrix of T.

Example 4The transformation defines a reflection in the line y = x.

It can be shown that T is linear. Determine the standard matrix of this

transformation. Find the image of .

y

yxyx

T3

2)(

Solution

We find the effect of T on the standard basis.

10

)01

(T and01

)10

(T0110

A

T can be written as

yx

yx

T0110

)(

14

41

14

0110

)14

(T

Figure V.6

End of Lecture