Ch. 5 Linear Models & Matrix Algebra 5.1Conditions for Nonsingularity of a Matrix 5.2Test 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

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)

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)

x + y = 9

x + y = 8

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

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

5.15.1 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)

1111120

3 & 1 rows swap

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)

3 & 1 rows swap

216131

3*216131

7*216131

31*23211

operations row Elementary

61014-

1A of Inverse

61014-

A Adjoint

2-14-6

A of Cofactors

6A oft Determinan

216131

3*216131

7*216131

31*23211

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)

65z 3;1y ;21

61014-

InversionMatrix

;5 ;2 ;3

Rule sCramer'

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

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

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

312213332112233211

322113312312332211

333231

232221

131211

scalaraaaaaa

aaA 12212211

scalarCaAn

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

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

222113

232112

232211

333231

232221

131211

ij MC 1

jjj CaA

5.2 Determinants5.2 Determinants

Pattern of the signs for cofactor minors

ij MC 1

61014-

1A of Inverse

61014-

A Adjoint

2-14-6

A of Cofactors

6A oft Determinan

216131

3*216131

7*216131

31*23211

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)

65z 3;1y ;21

61014-

InversionMatrix

;5 ;2 ;3

Rule sCramer'

5.2 Evaluating a 5.2 Evaluating a determinantdeterminant

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

0214827151271268859

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

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)

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

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

• 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

bTaIgG

bTagbIC

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)

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

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

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

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

ACAC''

Matrix ACMatrix AC''

CaCaCa

Inverse of AInverse of A

nIACA IAA

CAA 11

Solving for X using Matrix Solving for X using Matrix InversionInversion

adjAx *

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

ccP oo

Cramer’s ruleCramer’s rule

)(1100

AYbTaIbTa

bTagbI

ACbTagbI

AGIbTag

bTaIbTa

bTagbI

bTaIbTa

bTagbI

Deriving Cramer’s RuleDeriving Cramer’s Rule

CdCdCd

2222121

1212111

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)

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

jjj CaA

dAadjA

dAx 11

Finding the DeterminantFinding the Determinant

1Y – 1C–1G = I0

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

-gY+0C+ 1G = 0

Y = C+I0+G

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

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

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)

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

jjj CaA

dAadjA

dAx 11

5.4 Inverse, an example5.4 Inverse, an example

122121

Finding the DeterminantFinding the Determinant

1Y – 1C–1G = I0

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

-gY+0C+ 1G = 0

Y = C+I0+G

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

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

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

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

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

bTaIgG

bTagbIC

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)

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

dxaxaxax

222221212

112121111

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

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

ToSector 1

(a1jx1)Sector 2

(a2jx2)Final

demand(di)

Total output

Sector 1 150 500 350 1000

Sector 2 200 100 1700 2000

Factor Payment

650 1400 1100 3150

Total outlays

1000 2000 3150 6150

ToSector 1

(aij)Sector 2

(aij)Final

demand(di)

Total output

Sector 1 0.15 0.25 350 1000

Sector 2 0.20 0.05 1700 2000

Factor Payment

0.65 0.70 1100 3150

Total outlays

1.00 1.00 3150 6150

Leontief Input-output Leontief Input-output AnalysisAnalysis

05.20.

25.15.

05.20.

25.15.

05.20.

25.15.

05.20.

25.15.

05.120.

25.15.1

95.20.

25.85.

85.20.

25.95.

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

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)

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

Ch. 5 Linear Models & Matrix Algebra 5.1Conditions for Nonsingularity of a Matrix 5.2Test of...

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