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Ch. 5 Linear Models & Matrix Algebra. 5.1 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
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Ch. 5 Linear Models & Matrix Algebra 5.1 Conditions for Nonsingularity of a Matrix5.2 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
5.1 Conditions for Nonsingularity of a Matrix3.4 Solution of a General-equilibrium System (p. 44) • x + y = 8x + y = 9(inconsistent & dependent) • 2x + y = 124x + 2y= 24(dependent) • 2x + 3y = 58y = 18x + y = 20(over identified & dependent)
y 12 For both the equations Slope is -1 y x + y = 9 x + y = 8 x 5.1 Conditions for Non-singularity of a Matrix3.4 Solution of a General-equilibrium System (p. 44) • Sometimes equations are not consistent, and they produce two parallel lines. (contradict) • Sometimes one equation is a multiple of the other. (redundant)
5.1 Conditions for Non-singularity of a MatrixNecessary versus sufficient conditionsConditions for non-singularityRank 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 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 row These 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
5.1 Conditions for Nonsingularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 86)
5.1 Conditions for Non-singularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)
5.1 Conditions for Nonsingularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)
5.1 Conditions for Non-singularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)
5.2 Test of Non-singularity by Use of DeterminantDeterminants and non-singularityEvaluating a third-order determinantEvaluating 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
5.2 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 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
5.2 Test of Non-singularity by Use of Determinant and permutations: 2x2 and 3x3
5.2 Test of Non-singularity by Use of Determinant : 4 x 4 permutations = 24
5.2 Evaluating a third-order determinantEvaluating an 3 order determent by Laplace expansion Laplace Expansion by cofactors; if /A/ = 0, then /A/ is singular, i.e., under identified
5.2 Determinants • Pattern of the signs for cofactor minors
5.1 Conditions for Nonsingularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)
5.1 Conditions for Non-singularity of a MatrixConditions for non-singularity, Rank of a matrix (p. 96)
5.2 Evaluating a determinant • Laplace expansion of a 3rd order determinant by cofactors. If /A/ = 0, then singular
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
5.3 Basic Properties of DeterminantsProperties I to III (related to elementary row operations) • The interchange of any two rows will alter the sign but not its numerical value • The multiplication of any one row by a scalar k will change its value k-fold • The addition of a multiple of any row to another row will leave it unaltered.
5.3 Basic Properties of DeterminantsProperties IV to VI • The interchange of rows and columns does not affect its value • If one row is a multiple of another row, the determinant is zero • The expansion of a determinant by alien cofactors produces a result of zero
5.3 Basic Properties of DeterminantsProperties I to V • If /A/ 0 • Then • A is nonsingular • A-1 exists • A unique solution to • X=A-1d exists • /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
5.4 Finding the Inverse 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
A= C= C’=
5.4 Finding the Inverse MatrixExpansion of a determinant by alien cofactors, Property VI, Matrix inversion • Expansion by alien cofactors yields /A/=0 • This property of determinants is important when defining the inverse (A-1)
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-1A = I • We are interested in A-1 because x=A-1d
5.4 Finding the Inverse 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
Derivation of matrix inverse formula • |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 Finding the Inverse 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
5.4 A Inverse • A(adjA) = |A|I • A(adjA)/|A| = I ( |A| is a scalar) • A-1A(adjA)/|A|= A-1I • adjA/|A|= A-1
Finding 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
Derivation of matrix inverse formula • |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 Finding the Inverse 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
5.4 A Inverse • A(adjA) = |A|I • A(adjA)/|A| = I ( |A| is a scalar) • A-1A(adjA)/|A|= A-1I • adjA/|A|= A-1
Finding 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
= 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