Chiang - Chapter 5

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Notes fromCHAPTER 5 Alpha Chiang

LINEAR MODELS AND MATRIX ALGEBRA

5.1 Conditions for Nonsingularity of a MatrixWhen squareness condition is already met, a sufficient condition for thenonsingularity of a matrix is that its rows be linearly independent.

Squareness and linear independence constitute the necessary and sufficientcondition for non-singularity.

Nonsingularity squareness and linear independence

An n x n coefficient matrix A can be considered as an ordered set of row vectors,ie., as a column vector whose elements are themselves row vectors:

11 12 1 1

21 22 2 2

1 1 1

'

'

'

n

n

n n n n

a a a v

a a a vA

a a a v

= =

L

L

L L L L M

L

where [ ]1 2' , 1,2, ,i i i inv a a a i n= =L L

For the rows to be linearly independent, none must be a linear combination of therest.

Example:

'

1

'

2

'

3

3 4 5

0 1 2

6 8 10

v

A v

v

= =

Since [6 8 10] = 2 [3 4 5],

We have v3 = 2 v1 +2v1+0 v2. Thus the third row is expressible as a linearcombination of the first two, the rows are not linearly independent.

Rank of a Matrix

Rankr= the maximum number of linearly independent rows that can be found in amatrix. (it also tells us the maximum number of linearly independent columns inthe said matrix).

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By definition, an n x n nonsingular matrix has n linearly independent rows (orcolumns); consequently, it must be of rank n. Conversely, an n x n matrix have arank n must be nonsingular.

TEST OF NONSINGULARITY BY USE OF DETERMINANT

Determinants and Non-singularity

The determinant of a square matrix A, denoted by |A|, is a uniquely defined scalar(number) associated with that matrix.

Determinants are uniquely defined only for square matrices.

For a 2 x 2 matrix11 12

21 22

a aA

a a

=

, its determinant is defined to be the sum of two terms

as follows:

11 12

11 22 21 12

21 22

a a A a a a a

a a= = [= a scalar]

Example:

10 4 3 5

8 5 0 1Given A and B

= =

their determinants are10 410(5) 8(4) 18

8 5

3 53( 1) 0(5) 3

0 1

A

B

= = =

= = =

Relationship between linear dependence of rows in matrix A vs. determinant

1 1

2 2

' '3 8 2 6

' '3 8 8 24

c dC and D

c d

= = = =

both have linearly dependent rows because c1=c2 and d2=4d1.

Determinants:

3 83(8) 3(8) 0

3 8

2 62(24) 8(6) 0

8 24

C

D

= = =

= = =

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The result suggest that a vanishing determinant may have something to do with lineardependence.

Evaluating a Third Order Determinant

11 12 13

22 23 21 23 21 22

21 22 23 11 12 13

32 33 31 33 31 32

31 32 33

11 22 33 11 23 32 12 23 31 12 21 33 13 21 32 13 22 31

a a aa a a a a a

A a a a a a aa a a a a a

a a a

a a a a a a a a a a a a a a a a a a

= = +

= + +

a scalar

Examples

2 1 34 5 6 9

7 8 9

=

7 0 3

9 1 4 295

0 6 5

=

Evaluating nth order Determinant by Laplace Expansion

To sum up, the value of a determinant |A| of order n can be found by the Laplaceexpansion of any row or any column.

(5.8)1

n

ij ij

j

A a C =

= [expansion by ith row]

1

n

ij ij

i

A a C =

= [expansion by jth row]

5.3 BASIC PROPERTIES OF DETERMINANTS

Property I: The interchange of rows and columns does not affect the value of adeterminant.

Property II: The interchange of any two rows (or any two columns) will alter the sign,but not the numerical value of the determinant.

Property III. The multiplication of any one row (or one column) by a scalar k willchange the value of the determinant k-fold.

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( )ka kb a b

kad kbc k ad bc k c d c d

= = =

15 7 5 7 5 73 3(2) 6(5 14 )

12 2 4 2 2

a b a b a bad bc

c d c d c d

= = =

Property IV. The addition (subtraction) of a multiple of any row to (from) another rowwill leave the value of the determinant unaltered. The same holds true if we replace theword row by column in the above statement.

( ) ( )a b a b

a d kb b c ka ad bcc ka d kb c d

= + + = =+ +

Property V. If one row (or column) is a multiple of another row (or column) the valueof the determinant will be zero. As a special case, when two rows(or columns) areidentical, the determinant will vanish.

2 22 2 0 0

a b c cab ab cd cd

a b d d = = = =

Determinantal Criterion for Nonsingularity

Summary: Given a linear equation system Ax=d where a is an n x n coefficient matrix,

0A there is row (column) independence in matrix A

A is non-singular A-1 exists a unique solution 1x A d = exists

Caveat: we may sometimes get negative solution values that are economicallyinadmissible.

Example:

Does the equation system possess a unique solution?

1 2 3

1 2 3

2 3

7 3 3 7

2 4 0

2 2

x x x

x x x

x x

=

+ + =

=

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7 3 3

2 4 1 8 0.

0 2 1

=

Therefore a unique solution exists.

Rank of a Matrix Redefined

Rank of matrix A is defined as the maximum number of linearly independent rows inA. In view of the link between row independence and the nonvanishing of thedeterminant, we can redefine the rank of an m x n matrix as the maximum order of nonvanishing determinant that can be constructed from the rows and columns of thatmatrix.

r (A) min {m,n}

The rank of A is less than or equal to the minimum of the set of two numbers m and n.

5.4 FINDING THE INVERSE MATRIX

Expansion of a determinant by Alien Co-factors

Property VI. The expansion of a determinant by alien cofactors (the cofactors of awrong row or column) always yields a value of zero.

Example: If we expand the determinant

4 1 2

5 2 1

1 0 3

by using its first row elements but

the cofactors of the second row elements |c21|, |c22|, |c23|we get a11|c21| + a12 |c22| + a13 |c23| = 4(-3)+1(10)+2(1)

More generally, if we have a determinant

11 12 13

21 22 23

31 32 33

a a a

A a a a

a a a

= , will yield a zero sum as follows:

3

1 2 11 21 12 22 13 23

1

12 13 11 13 11 12

11 12 13

32 33 31 33 31 32

11 12 33 11 13 32 11 12 33 12 13 31 11 13 32 12 13 31 0

j j

i

a c a c a c a c

a a a a a aa a a

a a a a a a

a a a a a a a a a a a a a a a a a a

=

= + + =

= +

= + + + =

5.9

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Reason for this result: sum of products in 5.9 can be considered as the result of theregular expansion by the second order row of another determinant

11 12 13

11 12 13

31 32 33

*

a a a

A a a a

a a a

= which only differs from |A| only in its second row and whose first

two rows are identical.

(5.10) '1

0 ( ')n

ij i j

j

a C i i=

= [expansion by ith row and cofactors of i th

column]

'

1

0 ( ')n

ij i j

j

a C i i=

= [expansion by jth row and cofactors of jth

column]

Matrix Inversion

11 12 1

21 22 2

( )

1 2

( 0)

n

n

nxn

n n nn

a a a

a a aA A

a a a

=

L

L

L L L L

L

Form a matrix of cofactors by replacing each element aij with its cofactor |Cij|. Suchcofactor matrix denoted by C = [|Cij|] must also be n x n.

Our interest is the transpose C commonly referred to as adjoint of A.

11 21 1

12 22 2

( )

1 2

'

n

n

nxn

n n nn

C C C

C C CC adjA

C C C

L

L

L L L L

L

The matrices A and C are conformable for multiplication and their product AC isanother matrix n x n matrix in which each element is the sum of products.

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1 1 1 2 1

1 1 1

2 1 2 2 2

1 1 1( )

1 2 1

1 1 1

'

0 0 0 1 0 0 0

0 0 0 0 1 0 0

0 0 0 0 0 1 0

0 0 0 0 0 0 1

n n n

j j j j j nj

j j j

n n n

j j j j j nj

j j jnxn

n n n

nj j nj j j nj

j j j

a C a C a C

a C a C a C AC

a C a C a C

A

A A A I

A

A

= = =

= = =

= = =

=

= = =

L

L

M M M

L

n

by 5.8 and 5.10

As the determinant |A| is a non zero scalar, it is permissible to divide both sides of theequation AC = |A| I.

The result is' ' AC C

I or A I A A

= =

Premultiplying both sides of the last equation by A-1, and using the result that A-1A=I,

we can get1 1' 1

,

C

A or A adj AA A

= =

This is one way to invert matrix A!!!

Example:

11 12

21 22

1

312 2

3 22 0 inverse exists

1 0

0 12 3

0 2'

1 3

0 10 21 12 1 3

A A

C CCC C

C adjA

A adj AA

= =

= =

= =

= = =

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Example:

Find the inverse of

4 1 1

0 3 2

3 0 7

B

=

. Since |B|=99 0, the inverse of B-1 exists. The

cofactor matrix is3 2 0 2 0 3

0 7 3 7 3 021 6 9

1 1 4 1 4 17 31 3

0 7 3 7 3 05 8 12

1 1 4 1 4 1

3 2 0 2 0 3

=

Therefore,21 7 6

6 31 8

9 3 12

adj B

=

and the desired inverse matrix is

1

21 7 51 1

6 31 899

9 3 12

B adj BB

= =

Check if AA-1

= A-1

A = I and BB-1

= B-1

B = I.

5.5 CRAMERS RULE

- a practical way of solving a linear equation system.

- Given an equation system Ax d = where A is nxn , the solution can be written

as

1 1 ( ) x A d adj A d A

= =

provided A is non-singular.

[p.108 provides the proof is which rather long]

1 1

1x A

A=

|A1| is a new determinant were we replace the first column of |A| by the columnvector d but keep all the other columns intact.

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The expansion of the |A1| by its first column (the d column) will yield the

expression 11

n

i i

i

d C=

because the elements di now take the place of elements aij.

1 1

1x A

A

=

In general,

(5.15)

11 12 1 1

21 22 2 2

1 2

1

n

j n

j

n n n nn

a a d a

A a a d ax

A A

a a d a

= =

L L

L L

M M M

L L

Examples:

Find the solution of1 2

1 2

5 3 30

6 2 8

x x

x x

+ =

=

1 2

1

1

2

1

5 3 30 3 5 3028 84 140

6 2 8 2 6 8

843

28

140528

A A A

Ax

A

Ax A

= = = = = =

= = =

= = =

Find the solution of the equation system:

1 2 3

1 2 3

1 2 3

1 2 3

7 0

10 2 8

6 3 2 7

61, 61, 183, 244,

x x x

x x x

x x x

A A A A

=

+ =

+ =

= = = =

Work this out!!!!

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This is thestatement ofCramersRule


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Solutions:

1

1

2

2

2

3

611

61

183 361

2444

61

Ax

A

AxA

Ax

A

= = =

= = =

= = =

Note that |A| 0 is necessary condition for the application of Cramers Rule.Cramers rule is based upon the concept of the inverse matrix, even though in practice it

bypasses the process of matrix inversion.

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Chiang - Chapter 5