chiang ch5.ppt
TRANSCRIPT
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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
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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
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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
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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
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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
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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
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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
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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
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5.1 Conditions for Non-singularity of a Matrix Conditions for non-singularity, Rank of a matrix (p. 96)
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
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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
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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
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5.2 Test of Non-singularity by Use of
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
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5.2 Test of Non-singularity by Use of
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
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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
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5.2 Determinants
Pattern of the signs for cofactor minors
15
ij
ji
ij M C
1
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5.1 Conditions for Non-singularity of a Matrix Conditions for non-singularity, Rank of a matrix (p. 96)
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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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5.4 matrix A: matrix of parameters from
the equation Ax=d
nnnn
n
n
nxn
aaa
aaa
aaa
A
21
22221
11211
27
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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
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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
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A
A A
C A
00
0000
n I A A
100
010001
32
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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 *
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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
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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
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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
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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
40
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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
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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
42
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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
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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
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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
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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
46
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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
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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
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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
=
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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
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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
=
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3 4 S l i f G l E
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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
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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
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d xa xa xa x
d xa xa xa x
d xa xa xa x
2211
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(I -A)x = d ; x = (I -A)-1 d
55
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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
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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
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58
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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
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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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