Chiang Ch5.Ppt

7/27/2019 Chiang Ch5.Ppt

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Ch. 5 Linear Models & Matrix

Algebra5.1 Conditions for Nonsingularity of a Matrix

5.2 Test of Nonsingularity by Use of Determinant

5.3 Basic Properties of Determinants

5.4 Finding the Inverse Matrix

5.5 Cramer's Rule

5.6 Application to Market and National-Income Models

5.7 Leontief Input-Output Models 5.8 Limitations of Static Analysis

1



5.1 Conditions for Nonsingularity of a Matrix

3.4 Solution of a General-equilibrium System (p. 44)

x + y = 8

x + y = 9

(inconsistent &

dependent)

2x + y = 12

4x + 2y= 24

(dependent)

2x + 3y = 58

y = 18

x + y = 20

(over identified &

dependent)

9

8

11

11

y

x

24

12

24

12

y

x

2

20

1858

11

1032

y x



5.1 Conditions for Non-singularity of a Matrix

3.4 Solution of a General-equilibrium System (p. 44)

Sometimes

equations are not

consistent, and theyproduce two parallel

lines. (contradict)

Sometimes oneequation is a

multiple of the other.

(redundant) 3

y

x

x + y = 9

x + y = 8

y

12

For both theequations

Slope is -1



5.1 Conditions for Non-singularity of a Matrix Necessary versus sufficient conditions

Conditions for non-singularity

Rank of a matrix

A) Square matrix , i.e., n. equations = n.

unknowns. Then we may haveunique 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 Operations (p.

86)

Interchange any two rows in a matrix

Multiply 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



5.1 Conditions for Nonsingularity of a Matrix 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&1rowsswap100010

001

014262

4110



5.1 Conditions for Non-singularity of a Matrix 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&1rowsswap

100

010001

301

125214

21613113537

21211

100010

001

3*216131

13537

100

100

010

301

7*216131

25210

100

100

710

301

31*23211

25210

100

300

710

301

operationsrowElementary



5.1 Conditions for Nonsingularity of a Matrix Conditions for non-singularity, Rank of a matrix (p. 96)

8

312-

61014-

3-3-6

6

1Aof Inverse

312-

61014-

3-3-6

AAdjoint

363-

1103-2-14-6

Aof Cofactors

6Aof tDeterminan

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




9

65z3;1y;21

3

44

312-

61014-3-3-6

6

1

z

yx

312-

61014-

3-3-6

6

1A

3

4

4

d;

z

y

x

x;

301

125

214

InversionMatrix

1-

1

x

d A x

A

d Ax

65

31

21

;5;2;3

;6

3

4

4

d;

z

y

x

x;

301

125

214

RulesCramer'

3

2

1

321

A A z

A A y

A A x

A A A

A

A

d Ax



5.2 Test of Non-singularity by Use of Determinant Determinants and non-singularity

Evaluating a third-order determinant

Evaluating 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 possibleproducts

t(-1)t a1j a2k…ang, where the series of

second subscripts is a permutation of (1,..,

n) including the natural order (1, …, n), andt 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 10



5.2 Test of Non-singularity by Use of

Determinant P(n,r) = n!/(n-r)! P(2,2) = 2!/(2-2)! = 2

There are only two ways of arrangingsubscripts (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) = 2

3!=(3)(2)(1) = 64!=(4)(3)(2)(1) = 245!=(5)(4)(3)(2)(1) = 1206!=(6)(4)(3)(2)(1) = 720… …

10! =3,628,800 11




Determinant and permutations: 2x2 and 3x3

scalar aaaaaaaaa

aaaaaaaaa

aaa

aaa

aaa

A

312213332112233211

322113312312332211

333231

232221

13

1211

scalar aaaaaa

aa A 12212211

2221

1211

scalar C a An

j

j j 1

21

12




Determinant : 4 x 4 permutations = 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 determinant

Evaluating an 3 order determent by Laplace expansion

Laplace Expansion by cofactors;if /A/ = 0, then /A/ is singular, i.e., under identified

3231

2221

13

3331

2321

12

3332

2322

11

aa

aa M

aa

aa M

aa

aa

M

14

333231

232221

131211

aaaaaa

aaa

A

ij

ji

ij M C

1

n

j j j C a A 111



5.2 Determinants

Pattern of the signs for cofactor minors

15

ij

ji

ij M C

1






17

65z3;1y;21

3

44

312-

61014-3-3-6

6

1

z

yx

312-

61014-

3-3-6

6

1A

3

4

4

d;

z

y

x

x;

301

125

214

InversionMatrix

1-

1

x

d A x

A

d Ax

65

31

21

;5;2;3

;6

3

4

4

d;

z

y

x

x;

301

125

214

RulesCramer'

3

2

1

321

A A z

A A y

A A x

A A A

A

A

d Ax



5.2 Evaluating a

determinant 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 Use

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)2

Expansion 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 Determinants Properties I to III (related to elementary row operations)

20

I. The interchange of any two rows will

alter the sign but not its numerical

valueII. The multiplication of any one row by

a scalar k will change its value k-fold

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



5.3 Basic Properties of Determinants Properties 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 Determinants Properties 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 Inverse

aka “the hard way”

Steps in computing the Inverse Matrix andsolving for x

1. Find the determinant |A| using expansion bycofactors.

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



g b

bT a I g G

g b

bT a g bI C

g bbT a I Y

1

1

1

1

00

00

00

g b

g

b A

1

10

01

111

24

10

01

111

g

b A=

C=

bb

g g

g b

11

11

1

C’=

b g g

b g b

1

1

111

G

C

Y

0

0

0

bT a

I

b g g

b g b

g b 1

1

111

1

1



5.4 Finding the Inverse Matrix Expansion of a determinant by alien cofactors,

Property VI, Matrix inversion

Expansion by alien

cofactors yields

/A/=0

This property of

determinants isimportant when

defining the inverse

(A-1)25

01

21

n

j

j j C a A



5.4 A Inverse (A-1)

Inverse of A is A-1

if and only if A is square (nxn) and

rank = n AA-1 = A-1 A = I

We are interested in

A-1 because x=A-1d

26



5.4 matrix A: matrix of parameters from

the equation Ax=d

nnnn

n

n

nxn

aaa

aaa

aaa

A

21

22221

11211

27





C' or adjoint A: Transpose matrix of

the cofactors of A

Aadj

C C C

C C C

C C C

C

nnnn

n

n

nxn

21

22212

12111

29





Matrix AC'

n

j

njnj

n

j

jnj

n

j

jnj

n

jnj j

n

j j j

n

j j j

n

j

nj j

n

j

j j

n

j

j j

nxn

C aC aC a

C aC aC a

C aC aC a

C A

11

2

1

1

12

122

112

1

1

1

21

1

11

31



A

A A

C A

00

0000

n I A A

100

010001

32



Inverse of A

n I AC A I A A

C A A 11

I AC A

I A

C A

33

1 A AC

d A

C x *



Solving for X using Matrix

Inversion

d A

adjA x *

nnnnn

n

n

n d

d

d

C C C

C C C

C C C

A

x

x

x

2

1

21

22212

12111

2

1

1

34







37

)(1

100

01

11

00*

000

0

g b

bT a I

A

AY

bT a I bT a

I

A

Y

Y

)(1

1

1

10

0

11

00*

000

0

g b

bT a g bI

A

AC

bT a g bI

g

bT ab

I

A

C

C

)(1

00

1

11

00*

000

0

g b

I bT a g

A

AG

I bT a g

g

bT ab

I

A

G

G



38

000

0

100

01

11

bT a I bT a

I

AY

000

0

1100

11

bT a g bI g

bT ab

I

AC

000

0

00

1

11

bT a I g

g

bT ab

I

AG










cofactor matrix.




40



Derivation of matrix inverse formula

|A| = ai1ci1 + …. + aincin (scalar)

Adj. A = transposed cofactor matrix of A

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

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

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

41








cofactor matrix.




42



5.4 A Inverse

A(adj A) = |A|I

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

A-1 A(adj A)/|A|= A-

1I

adj A/|A|= A-1

43

n

j

j j C a A1

11

d Aadj A

d A x 11



Finding 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

g b

g b g b

g

b D

1

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

10

01

111



Derivation of matrix inverse formula

|A| = ai1ci1 + …. + aincin (scalar)

Adj. A = transposed cofactor matrix of A

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

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

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

45








cofactor matrix.




46





5.4 Inverse, an example

o

o

P P

c P c P c

2211

2211

48

o

oc

P

P cc

2

1

21

21

1221

21

21

cc

cc A

12

12

ccC

11

22

c

cadjA

o

occc

cc P P

11

22

12212

1 1



Finding 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

g b

g b g b

g

b D

1

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

10

01

111



The 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

bT a

I

=



Macro model

Section 3.5, Exercise 3.5-2 (a-d), p. 47

Section 5.6, Exercise 5.6-2 (a-b), p. 111

Given 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

rule51



The 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

bT a

I

=



3 4 S l i f G l E



3.4 Solution of General Eq.

System

(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

2058

1132

y x



5.7 Leontief Input-Output Models Structure of an input-output model

The open model, A numerical example

Finding the inverse by approximation, The closed model

nnnnnnn

nn

nn

d xa xa xa x

d xa xa xa x

d xa xa xa x

2211

222221212

112121111

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

55

3

2

1

2

1

21

22221

11211

11

1

d d

d

x

x

x

aaaaaa

aaa

n

nnnn

n

n



56

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

To Sector 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 (W i

) 650 1400 1100 3150

Total outlays

(X i ) 1000 2000 3150 6150

To Sector 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 (W i ) 0.65 0.70 1100 3150

Total outlaysX

1.00 1.00 3150 6150



Leontief Input-output Analysis

iij

iiij

iiiji

iijii

iiiij

iiiij

d A I x

d x A I

d x A x I

x A x I d

x I d x A

xd x A

i

1*

57

21 d



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



5.8 Limitations of Static Analysis

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 theequilibrium

59

3 4 S l ti f G l E



3.4 Solution of General Eq.

System

(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

2058

1132

y x



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Chiang Ch5.Ppt