Ch. 5 Linear Models & Matrix Algebra 5.1Conditions for Nonsingularity of a Matrix 5.2Test of Nonsingularity by Use of Determinant 5.3Basic Properties of

Ch. 5 Linear Models & Matrix Ch. 5 Linear Models & Matrix Algebra Algebra

5.1 Conditions for Nonsingularity of a Matrix5.2 Test of Nonsingularity by Use of Determinant5.3 Basic Properties of Determinants5.4 Finding the Inverse Matrix5.5 Cramer's Rule5.6 Application to Market and National-Income Models5.7 Leontief Input-Output Models5.8 Limitations of Static Analysis

1

5.15.1 Conditions for Nonsingularity of a MatrixConditions for Nonsingularity of a Matrix3.4 3.4 Solution of a General-equilibrium System Solution of a General-equilibrium System (p. 44)(p. 44)

x + y = 8x + y = 9(inconsistent & dependent)

2x + y = 124x + 2y= 24(dependent)

2x + 3y = 58y = 18x + y = 20(over identified & dependent)

9

8

11

11

y

x

24

12

24

12

y

x

2

20

18

58

11

10

32

y

x

5.15.1 Conditions for Non-singularity of a Conditions for Non-singularity of a MatrixMatrix3.4 3.4 Solution of a General-equilibrium System Solution of a General-equilibrium System (p. 44)(p. 44)

Sometimes equations are not consistent, and they produce two parallel lines. (contradict)

Sometimes one equation is a multiple of the other. (redundant)

3

y

x

x + y = 9

x + y = 8

y

12

For both the equations

Slope is -1

5.1 Conditions for Non-singularity of a Matrix5.1 Conditions for Non-singularity of a MatrixNecessary versus sufficient conditionsNecessary versus sufficient conditionsConditions for non-singularityConditions for non-singularityRank of a matrixRank of a matrix

A) Square matrix , i.e., n. equations = n. unknowns. Then we may have unique solution. (nxn , necessary)

B) Rows (cols.) linearly independent (rank=n, sufficient)

A & B (nxn, rank=n) (necessary & sufficient), then nonsingular

4

5.1 Elementary Row Operations5.1 Elementary Row Operations (p. 86)(p. 86)

Interchange any two rows in a matrixMultiply or divide any row by a scalar k (k

0)Addition of k times any row to another rowThese operations will:

◦transform a matrix into a reduced echelon matrix (or identity matrix if possible)

◦not alter the rank of the matrix◦place all non-zero rows before the zero

rows in which non-zero rows reveal the rank

5

5.15.1 Conditions for Nonsingularity of a MatrixConditions for Nonsingularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 86)Conditions for non-singularity, Rank of a matrix (p. 86)

6

121

1111120

4100

000

11410

0411

11*

001

1111120

4100

4110

11410

0411

112*

001

2110

4100

4110

22110

0411

2*

001

010

4100

4110

262

0411

41*

001

010

100

4110

262

014

3 & 1 rows swap

100

010

001

014

262

4110

5.15.1 Conditions for Non-singularity of a MatrixConditions for Non-singularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)Conditions for non-singularity, Rank of a matrix (p. 96)

7

1*

401

25210

100

1010

710

301

4*

001

25210

100

214

710

301

21*

001

510

100

214

1420

301

5*

001

010

100

214

125

301

3 & 1 rows swap

100

010

001

301

125

214

216131

13537

21211

100

010

001

3*216131

13537

100

100

010

301

7*216131

25210

100

100

710

301

31*23211

25210

100

300

710

301

operations row Elementary


8

312-

61014-

3-3-6

6

1A of Inverse

312-

61014-

3-3-6

A Adjoint

363-

1103-

2-14-6

A of Cofactors

6A oft Determinan

301

125

214

A

216131

13537

21211

100

010

001

3*216131

13537

100

100

010

301

7*216131

25210

100

100

710

301

31*23211

25210

100

300

710

301

5.15.1 Conditions for Non-singularity of a Conditions for Non-singularity of a MatrixMatrixConditions for non-singularity, Rank of a Conditions for non-singularity, Rank of a matrix (p. 96)matrix (p. 96)

9

65z 3;1y ;21

3

4

4

312-

61014-

3-3-6

6

1

z

y

x

312-

61014-

3-3-6

6

1A

3

4

4

d ;

z

y

x

x;

301

125

214

InversionMatrix

1-

1

x

dAx

A

dAx

65

31

21

;5 ;2 ;3

;6

3

4

4

d ;

z

y

x

x;

301

125

214

Rule sCramer'

3

2

1

321

AAz

AAy

AAx

AAA

A

A

dAx

5.2 Test of Non-singularity by Use of Determinant5.2 Test of Non-singularity by Use of DeterminantDeterminants and non-singularityDeterminants and non-singularityEvaluating a third-order determinantEvaluating a third-order determinantEvaluating an nth order determent by Laplace expansionEvaluating an nth order determent by Laplace expansion

Determinant |A| is a uniquely defined scalar associated w/ a square matrix A(Chiang & Wainwright, p. 88)

|A| defined as the sum of all possible products t(-1)t a1j a2k…ang, where the series of second subscripts is a permutation of (1,.., n) including the natural order (1, …, n), and t is the number of transpositions required to change a permutation back into the original order (Roberts & Schultz, p. 93-94)

t equals P(n,r)=n!/(n-r)!, i.e., the permutation of n objects taken r at a time

10

5.25.2 Test of Non-singularity by Use of Test of Non-singularity by Use of DeterminantDeterminant

P(n,r) = n!/(n-r)! P(2,2) = 2!/(2-2)! = 2There are only two ways of arranging subscripts

(i,k) of product (-1)ta1ja2k either (1,2) or (2,1)The first permutation is even & positive (-1)2

and second is odd and negative (-1)1

0!=(1) = 11!=(1) = 12!=(2)(1) = 23!=(3)(2)(1) = 64!=(4)(3)(2)(1) = 245!=(5)(4)(3)(2)(1) = 120 6!=(6)(4)(3)(2)(1) = 720… …10! =3,628,800

11

5.25.2 Test of Non-singularity by Use of Test of Non-singularity by Use of Determinant and permutations: 2x2 Determinant and permutations: 2x2 and 3x3 and 3x3

scalaraaaaaaaaa

aaaaaaaaa

aaa

aaa

aaa

A

312213332112233211

322113312312332211

333231

232221

131211

scalaraaaaaa

aaA 12212211

2221

1211

scalarCaAn

jjj

121

12

5.25.2 Test of Non-singularity by Use Test of Non-singularity by Use of Determinant : of Determinant : 4 x 4 permutations 4 x 4 permutations = 24= 24

Abcd Bacd Cabd Dabc

Abdc Badc Cadb Dacb

Acbd Bcad Cbad Dbac

Acdb Bcda Cbda Dbca

Adbc Bdac Cdab Dcab

Adcb Bdca Cdba Dcba

13

5.2 Evaluating a third-order determinant5.2 Evaluating a third-order determinantEvaluating an 3 order determent by Laplace Evaluating an 3 order determent by Laplace expansionexpansionLaplace Expansion by cofactors;

if /A/ = 0, then /A/ is singular, i.e., under identified

3231

222113

3331

232112

3332

232211

aa

aaM

aa

aaM

aa

aaM

14

333231

232221

131211

aaa

aaa

aaa

A

ijji

ij MC 1

n

jjj CaA

111

5.2 Determinants5.2 Determinants

Pattern of the signs for cofactor minors

15

ijji

ij MC 1


16

312-

61014-

3-3-6

6

1A of Inverse

312-

61014-

3-3-6

A Adjoint

363-

1103-

2-14-6

A of Cofactors

6A oft Determinan

301

125

214

A

216131

13537

21211

100

010

001

3*216131

13537

100

100

010

301

7*216131

25210

100

100

710

301

31*23211

25210

100

300

710

301

5.15.1 Conditions for Non-singularity of a Conditions for Non-singularity of a MatrixMatrixConditions for non-singularity, Rank of a Conditions for non-singularity, Rank of a matrix (p. 96)matrix (p. 96)

17

65z 3;1y ;21

3

4

4

312-

61014-

3-3-6

6

1

z

y

x

312-

61014-

3-3-6

6

1A

3

4

4

d ;

z

y

x

x;

301

125

214

InversionMatrix

1-

1

x

dAx

A

dAx

65

31

21

;5 ;2 ;3

;6

3

4

4

d ;

z

y

x

x;

301

125

214

Rule sCramer'

3

2

1

321

AAz

AAy

AAx

AAA

A

A

dAx

5.2 Evaluating a 5.2 Evaluating a determinantdeterminant

Laplace expansion of a 3rd order determinant by cofactors. If /A/ = 0, then singular

0214827151271268859

23

567

13

468

12

459

123

456

789

A

18

5.2 Test of Non-singularity by 5.2 Test of Non-singularity by Use of DeterminantUse of Determinant

P(3,3) = 3!/(3-3)! = 6 |A| = 1(5)9 + 2(6)7 + 3(8)4

-3(5)7 – 6(8)1 – 9(4)2Expansion by cofactors

|A|= (1)c11 + (2)c12 + (3)c13

C11 = 5(9) – 6(8)

C12 = -4(9) + 6(7)

C13 = 4(8) – 7(5)Expansion across any row or

column will give the same # for the determinant

19

987

654

321

A

5.3 Basic Properties of Determinants5.3 Basic Properties of DeterminantsProperties I to III (related to elementary row Properties I to III (related to elementary row operations)operations)

20

I. The interchange of any two rows will alter the sign but not its numerical value

II. The multiplication of any one row by a scalar k will change its value k-fold

III. The addition of a multiple of any row to another row will leave it unaltered.

5.3 Basic Properties of Determinants5.3 Basic Properties of DeterminantsProperties IV to VIProperties IV to VI

21

IV. The interchange of rows and columns does not affect its value

V. If one row is a multiple of another row, the determinant is zero

VI. The expansion of a determinant by alien cofactors produces a result of zero

5.3 Basic Properties of Determinants5.3 Basic Properties of DeterminantsProperties I to VProperties I to V

/A/ = /A'/ Changing rows or col.

does not change # but changes the sign of /A/

k(row) = k/A/ka ± row or col.b =/A/ If row or col a=kb, then

/A/ =0

22

• If /A/ 0• Then

– A is nonsingular

– A-1 exists

– A unique solution to

X=A-1d exists

5.4 Finding the Inverse5.4 Finding the Inverse aka “the hard way” aka “the hard way”

Steps in computing the Inverse Matrix and solving for x

1. Find the determinant |A| using expansion by cofactors. If |A| =0, the inverse does not exist.

2. Use cofactors from step 1 and complete the cofactor matrix.

3. Transpose the cofactor matrix => adjA 4. Divide adj.A by |A| => A-1

5. Post multiply matrix A-1 by column vector of constants d to solve for the vector of variables x

23

gb

bTaIgG

gb

bTagbIC

gb

bTaIY

1

1

1

1

00

00

00

gb

g

bA

1

10

01

111

24

10

01

111

g

bA=

C=

bb

gg

gb

11

11

1

C’=

bgg

bgb

1

1

111

G

C

Y

0

0

0

bTa

I

bgg

bgbgb

1

1

111

1

1

5.45.4 Finding the Inverse MatrixFinding the Inverse MatrixExpansion of a determinant by alien Expansion of a determinant by alien cofactors, Property VI, Matrix cofactors, Property VI, Matrix inversioninversion

Expansion by alien cofactors yields /A/=0

This property of determinants is important when defining the inverse (A-1)

25

01

21

n

jjj CaA

5.4 A Inverse (A5.4 A Inverse (A-1-1))

Inverse of A is A-1 if and only if A is square (nxn)

and rank = nAA-1 = A-1A = IWe are interested in

A-1 because x=A-1d

26

5.4 matrix A: matrix of parameters 5.4 matrix A: matrix of parameters from the equation Ax=dfrom the equation Ax=d

nnnn

n

n

nxn

aaa

aaa

aaa

A

21

22221

11211

27

C: Matrix of cofactors of AC: Matrix of cofactors of A

nnnn

n

n

CCC

CCC

CCC

C

21

22221

11211

28

CC' ' or adjoint Aor adjoint A: T: Transpose ranspose matrix of the cofactors of A matrix of the cofactors of A

Aadj

CCC

CCC

CCC

C

nnnn

n

n

nxn

21

22212

12111

29

ACAC''

nnnn

n

n

aaa

aaa

aaa

21

22221

11211

nnnn

n

n

CCC

CCC

CCC

21

22212

12111

30

Matrix ACMatrix AC''

n

jnjnj

n

jjnj

n

jjnj

n

jnjj

n

jjj

n

jjj

n

jnjj

n

jjj

n

jjj

nxn

CaCaCa

CaCaCa

CaCaCa

CA

112

11

12

122

112

11

121

111

31

A

A

A

CA

00

00

00

nIAA

100

010

001

32

Inverse of AInverse of A

nIACA IAA

CAA 11

IA

CA

IA

CA

33

1A

A

C

dA

Cx

*

Solving for X using Matrix Solving for X using Matrix InversionInversion

dA

adjAx *

nnnnn

n

n

n d

d

d

CCC

CCC

CCC

A

x

x

x

2

1

21

22212

12111

2

1

1

34

5.4 A Inverse, solving for 5.4 A Inverse, solving for PP

o

oc

c

c

ccP

P

11

22

12212

1 1

35

1221

221

cc

ccP oo

1221

01102

cc

ccP

Cramer’s ruleCramer’s rule

36

)(1100

01

1100*

000

0

gb

bTaI

A

AYbTaIbTa

I

A YY

)(1

11

10

0

1100*

000

0

gb

bTagbI

A

ACbTagbI

g

bTab

I

A CC

)(1

00

1

1100*

000

0

gb

IbTag

A

AGIbTag

g

bTab

I

A GG

37

)(1

100

01

11

00*

000

0

gb

bTaI

A

AY

bTaIbTa

I

A

Y

Y

)(1

1

1

10

0

11

00*

000

0

gb

bTagbI

A

AC

bTagbI

g

bTab

I

A

C

C

)(1

00

1

11

00*

000

0

gb

IbTag

A

AG

IbTag

g

bTab

I

A

G

G

38

000

0

100

01

11

bTaIbTa

I

AY

000

0

1

10

0

11

bTagbI

g

bTab

I

AC

000

0

00

1

11

bTaIg

g

bTab

I

AG

Deriving Cramer’s RuleDeriving Cramer’s Rule

nnnnn

nn

nn

n

CdCdCd

CdCdCd

CdCdCd

A

x

x

x

2211

2222121

1212111

2

1

1

39

n

iini

n

iii

n

iii

nCd

Cd

Cd

A

x

x

x

1

12

11

2

1

1







40

Derivation of matrix inverse Derivation of matrix inverse formulaformula

|A| = ai1ci1 + …. + aincin (scalar)

Adj. A = transposed cofactor matrix of A

A(adj.A)=|A|I (expansion by alien cofactors = 0 for off diagonal elements)

A(adj.A)/|A| = I

A-1 = (adj.A)/|A| QED Roberts & Schultz, p. 97-8)

41







42

5.4 A Inverse5.4 A Inverse

A(adjA) = |A|I

A(adjA)/|A| = I ( |A| is a scalar)

A-1A(adjA)/|A|= A-1I

adjA/|A|= A-1

43

n

jjj CaA

111

dAadjA

dAx 11

Finding the DeterminantFinding the Determinant

1Y – 1C–1G = I0

-bY+1C+ 0G = a-bT0

-gY+0C+ 1G = 0

44

Y = C+I0+G

C = a + b(Y-T0)G = gY

gb

gbgb

g

bD

1

)1)()(1()0)()(1()0)()(1())(1)(1()0)(0(1)1)(1(1

10

01

111

Derivation of matrix inverse Derivation of matrix inverse formulaformula

|A| = ai1ci1 + …. + aincin (scalar)

Adj. A = transposed cofactor matrix of A

A(adj.A)=|A|I (expansion by alien cofactors = 0 for off diagonal elements)

A(adj.A)/|A| = I

A-1 = (adj.A)/|A| QED Roberts & Schultz, p. 97-8)

45







46

5.4 A Inverse5.4 A Inverse

A(adjA) = |A|I

A(adjA)/|A| = I ( |A| is a scalar)

A-1A(adjA)/|A|= A-1I

adjA/|A|= A-1

47

n

jjj CaA

111

dAadjA

dAx 11

5.4 Inverse, an example5.4 Inverse, an example

o

o

PP

cPcPc

2211

2211

48

o

oc

P

Pcc

2

1

21

21

122121

21

cccc

A

12

12

ccC

11

22

c

cadjA

o

oc

c

c

ccP

P

11

22

12212

1 1

Finding the DeterminantFinding the Determinant

1Y – 1C–1G = I0

-bY+1C+ 0G = a-bT0

-gY+0C+ 1G = 0

49

Y = C+I0+G

C = a + b(Y-T0)G = gY

gb

gbgb

g

bD

1

)1)()(1()0)()(1()0)()(1())(1)(1()0)(0(1)1)(1(1

10

01

111

The macro modelThe macro model

Y=C+I0+G 1Y - 1C – 1G = I0

C=a+b*(Y-T0) -bY + 1C + 0G = a-bT0

G=g*Y -gY + 0C +1G = 0

50

10

01

111

g

b

G

C

Y

0

0

0

bTa

I

=

Macro modelMacro model

Section 3.5, Exercise 3.5-2 (a-d), p. 47Section 5.6, Exercise 5.6-2 (a-b), p. 111Given the following model

(a) Identify the endogenous variables(b) Give the economic meaning of the parameter g(c) Find the equilibrium national income

(substitution)(d) What restriction on the parameters is needed for

a solution to exist?Find Y, C, G by (a) matrix inversion (b) Cramer’s rule

51

The macro modelThe macro model

Y=C+I0+G 1Y - 1C – 1G = I0

C=a+b*(Y-T0) -bY + 1C + 0G = a-bT0

G=g*Y -gY + 0C +1G = 0

52

10

01

111

g

b

G

C

Y

0

0

0

bTa

I

=

gb

bTaIgG

gb

bTagbIC

gb

bTaIY

1

1

1

1

00

00

00

gb

g

bA

1

10

01

111

53

10

01

111

g

bA=

C=

bb

gg

gb

11

11

1

C’=

bgg

bgb

1

1

111

G

C

Y

0

0

0

bTa

I

bgg

bgbgb

1

1

111

1

1

3.4 Solution of General Eq. 3.4 Solution of General Eq. SystemSystem(1)(1)-(1)(1) = 0

(inconsistent & dependent)

(2)(2)-(1)(4) = 0(dependent)

(2)(1)-(1)(3) = -1(independent as rewritten)

9

8

11

11

y

x

24

12

24

12

y

x

54

20

58

11

32

y

x

5.7 Leontief Input-Output Models5.7 Leontief Input-Output ModelsStructure of an input-output modelStructure of an input-output modelThe open model, A numerical exampleThe open model, A numerical exampleFinding the inverse by approximation, The closed Finding the inverse by approximation, The closed

modelmodel

nnnnnnn

nn

nn

dxaxaxax

dxaxaxax

dxaxaxax

2211

222221212

112121111

(I -A)x = d ; x = (I -A)-1 d

55

3

2

12

1

21

22221

11211

1

1

1

d

d

d

x

x

x

aaa

aaa

aaa

nnnnn

n

n

56

Miller and Blair 2-3, Table 2-3, p 15 Economic Flows ($ millions)

ToSector 1

(a1jx1)Sector 2

(a2jx2)Final

demand(di)

Total output

(xi)

Sector 1 150 500 350 1000

Sector 2 200 100 1700 2000

Factor Payment

(Wi)

650 1400 1100 3150

Total outlays

(Xi)

1000 2000 3150 6150

ToSector 1

(aij)Sector 2

(aij)Final

demand(di)

Total output

(xi)

Sector 1 0.15 0.25 350 1000

Sector 2 0.20 0.05 1700 2000

Factor Payment

(Wi)

0.65 0.70 1100 3150

Total outlays

(Xi)

1.00 1.00 3150 6150

Leontief Input-output Leontief Input-output AnalysisAnalysis

iij

iiij

iiiji

iijii

iiiij

iiiij

dAIx

dxAI

dxAxI

xAxId

xIdxA

xdxA

i

1*

57

58

2

1

2

1

2

1

05.20.

25.15.

x

x

d

d

x

x

2

1

2

1

2

1

10

01

05.20.

25.15.

x

x

d

d

x

x

2

1

2

1

2

1

05.20.

25.15.

10

01

x

x

x

x

d

d

2

1

2

1

2

1

05.20.

25.15.

10

01

d

d

x

x

x

x

2

1

2

1

05.120.

25.15.1

d

d

x

x

2

1

1

2

1

95.20.

25.85.

d

d

x

x

2000

1000

1700

350

85.20.

25.95.

7575.

1

2

1

x

x

5.85.8 Limitations of Static Limitations of Static AnalysisAnalysis Static analysis solves for the

endogenous variables for one equilibrium

Comparative statics show the shifts between equilibriums

Dynamics analysis looks at the attainability and stability of the equilibrium

59

3.4 Solution of General Eq. 3.4 Solution of General Eq. SystemSystem(1)(1)-(1)(1) = 0

(inconsistent & dependent)

(2)(2)-(1)(4) = 0(dependent)

(2)(1)-(1)(3) = -1(independent as rewritten)

9

8

11

11

y

x

24

12

24

12

y

x

60

20

58

11

32

y

x

5.6 Application to Market and National-Income 5.6 Application to Market and National-Income ModelsModels

Market modelMarket modelNational-income modelNational-income modelMatrix algebra vs. elimination of variablesMatrix algebra vs. elimination of variables

Why use matrix method at all?Compact notationTest existence of a unique

solutionHandy solution expressions

subject to manipulation

61

Documents

Ch. 5 Linear Models & Matrix Algebra 5.1Conditions for Nonsingularity of a Matrix 5.2Test of Nonsingularity by Use of Determinant 5.3Basic Properties of