TOPIC III LINEAR ALGEBRA

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1 [1] Linear Equations TOPIC III LINEAR ALGEBRA (1) Case of Two Endogenous Variables 1) Linear vs. Nonlinear Equations Linear equation: ax + by = c, where a, b and c are constants. 2 Nonlinear equation: ax + by = c. Can express a linear equation by a line in a graph. 2 EX 1: -2x + y = 1 EX 2: -x + y = 0 (nonlinear equation) 2) System of Linear Equations m equations: a1x + b2y = c1 a2x + b2y = c2 : a x + b y = c m m m III-1

2 If ( x, ȳ) satisfies all of the equations, it is called a solution. 3 possible cases: Unique solution No solution Infinitely many solutions EX 1: Case of no solution (inconsistent equations) 1) x - y = 1; 2) x - y = 2 inconsistent. No solution. EX 2: Case of infinitely many solutions (two equations and one redundant equation) 1) x - y = 1; 2) 2x - 2y = 2 ( xȳ, ) = (1,0), (2,1),... EX 3: Case of unique solution (two equations and no inconsistent and no redundant equations) 1) x - y = 1; 2) x + y = 1 ( xȳ, ) = (1,0) (unique) (2) Extension to Systems of Multiple Endogenous Variables System of m equations and n endogenous variables: a11x 1 + a12x a1nx n = b1 a21x 1 + a22x a2nx n = b2 : a x + a x a x = b m1 1 m2 2 mn n m III-2

3 variables: x,..., x 1 n constants: a ij, b i (i = 1,...,m; j = 1,...n) If ( x 1,..., x n ) satisfies all of the equations, it is called a solution. 3 possible cases. Is there any systematic way to solve the system? III-3

4 [3] Matrix and Matrix Operations Definition: A matrix, A, is a rectangular array of numbers: a 11 a 12 a 1n A a 21 a 22 a 2n a m1 a m2 a mn [a ij ] m n where i denotes row and j denotes columns. A is called a m n matrix. (m = # of rows ; n = # of columns.) , [ ], [4] (scalar) Definition: If m = n for a m n matrix A, A is called a square matrix III-4

5 Definition: t Let A be a m n matrix. The transpose of A is denoted by A (or A), which is a n m matrix; and it is obtained by the following procedure. 1st column of A 1st row of A t t 2nd column of A 2st column of A... etc. A ;A Definition: t Let A be a square matrix. A is called symmetric if and only if A = A (or A). A A t. Note: t For any matrix A, A A is always symmetric. Definition: Let A = [a ij] be a m n matrix. If all of the a ij = 0, then A is call a zero matrix. III-5

6 A ;B A is not a zero matrix, but B is. Definition: Let A and B are m n matrices. A + B is obtained by adding corresponding entries of A and B ; Definition: Let A be a m n matrix and c be a scalar (real number). Then, ca is obtained by multiplying all the entries of A by c Definition: III-6

7 b 11 A 1 a 11 a 12 a 1p ; B 1 b 21 A B = a b + a b a b p p1 b p1 4 A ; B A B = = Definition: Let A and B are m p and p n matrices, respectively. a 11 a 12 a 1p A 1 b 11 b 12 b 1n Let A a 21 a 22 a 2p A 2 ;B b 21 b 22 b 2b B 1 B 2 B n. Then, a m1 a m2 a mp A m b p1 b p2 b pn A 1 B 1 A 1 B 2 A 1 B n AB A 2 B 1 A 2 B 2 A 2 B n A m B 1 A m B 2 A m B n m n III-7

8 EX 1: A ; B AB EX 2: a11x 1 + a12x a1nx n = b1 a21x 1 + a22x a2nx n = b2 : am1x 1 + am2x amnx n = bm Define: a 11 a 12 a 1n x 1 b 1 A a 21 a 22 a 2n x 2 ; x = ; b = : b 2 : a m1 a m2 a mn x n b m Observe that III-8

9 a 11 x 1 a 12 x 2 a 1n x n b 1 a 21 x 1 a 22 x 2 a 2n x n b 2 Ax = = = b. : a m1 x 1 a m2 x 2 a mn x n b m EX 3: Thus, we can express the above system of equations by: Ax = b. x 1 + 2x 2= 1; x 2= 0 1 x x 2 = 1 0 x + 1 x = x x 2 0 Some Caution in Matrix Operations: (1) AB may not equal BA A =, B = Can show AB BA (2) Suppose that AB = AC. It does not mean that B = C. (3) AB = 0 (zero matrix) does not mean that A = 0 or B = 0. III-9

10 A = ; B = ; C = ; D = Can show AB = AC and AD = 0 (zero matrix) 2 2 Definition: Let A be a square matrix. A is call an identity matrix if all of the diagonal entries are one and all of the off-diagonals are zero. I Note: For A, I A = A and A I = A. m n m m n m n m n n m n Definition: For A n n and B n n, B is the inverse of A iff AB = I n or BA = I n A = ; B = AB = B = A Note: The inverse is unique, if it exists. III-10

11 Theorem: a b 1 d b -1 A 2 2 = A =. If ad = bc, no inverse. c d ad bc c a Theorem: EX 1: EX 2: Suppose that A and B are invertible. Then, (AB) = B A. A x m n n 1 b m 1 n n n n n n n A = A A A, n times. (A ) = A A = (A ). A system of linear equations is given by Assume m = n, and A is invertible. Then, x AAx = Ab I x = A b x = : x n -1 = A b (solution). III-11

12 [4] Determinant Definition: a 11 a 12 Let A 2 2 =. Then, A det(a) = a11a 22 - a12a 21. a 21 a A = det(a) = 8-3 = Definition: a 11 a 12 a 13 A = a 21 a 22 a a 31 a 32 a 33 a 11 a 12 a 13 a 11 a 12 a 13 a 21 a 22 a 23 a 21 a 22 a 23 a 31 a 32 a 33 a 31 a 32 a 33 det(a) = a a a + a a a + a a a - a a a - a a a - a a a A = det(a) = = = III-12

13 Definition: Let A n n = [a ij]. Then, th Minor of a ( M ) = (n-1) (n-1) matrix excluding i row and j ij ij column of A n n i+j Cofactor of a ij ( C ij ) = (-1) M ij th Definition: (Laplace expansion) n j1 det(a) = a ij C ij for any given i = a C a C. i1 i1 in in n i1 = a ij C ij for any given j = a C a C. 1j 1j nj nj This holds for general n. A = st Choose 1 row: 5 1 a = 1: M C 11 = (-1) M 11 = M a 12 = 2 : C 12 = (-1) 15 = -15 III-13

14 4 5 a = 3 : M C 13 = (-1) 7 = 7 det(a) = a c + a c + a c = (-15) = nd Choose the 2 row and do Laplace expansion (Do it by yourself) A = det(a) = 3(-1) Theorem: th Consider A n n. If all of the entries in the i row of A are zero, det(a) = 0. Definition: (Triangular matrices) A = upper t.m; B = lower t.m (not triangular) III-14

15 B = Theorem: Let A be a triangular matrix. Then, det(a) = product of diagonals. n n det(a) = = 24; det(b) = = 18 Theorem: (a) If B is a matrix that results when a single row (column) of A is multiplied by a constant k, then det(b) = k det(a) A = ; B = det(b) = 2 det(a). something same as A Same results when A = something & B = same as A III-15

16 1 2 3 A = ; B = det(b) = 2 det = 2 2 det = 4 det(a) (b) If B results when a multiple of one row (column) of A is added to another row (column), det(b) = det(a) A = ; B = det(b) = det(a) something same as A Same results when A =... & B = (c) If B results when two rows (columns) of A are interchanged, det(b) = -det(a) III-16

17 A = ; B = det(b) = -det(a) same same Same results if A = 2 5 same & B = 5 2 same A = det(a) = det = Note: If two rows or two columns are the same, then det = 0. EX 1: A = det(a) = = st st (1) A = row of A = 4 1 row of A III-17

18 det(a ) = 4 det(a) = -8 [by (a)] nd nd st (2) A = row of A = 2 row of A row of A det(a ) = det(a) = -2 [by (b)] st nd (3) A = and 2 rows of A interchanged det(a ) = -det(a) = 2 [by (c)] 3 EX 2: A = III-18

19 det(a) = -det = (-)(-)det = (+) = EX 3: B = det = det = [2nd row - (3/2) 1st row] [3rd row - 2 1st row] [3rd row + (4/3) 2nd row] Definition: t Let A m n = [a ij]. Then, A (transpose of A) = [a ji] n m t A = A = III-19

20 Theorem: (a) (b) (c) (d) t t (A ) = A t t t (A+B) = A + B t t t (AB) = B A t det(a ) = det(a). Theorem: Let A and B be n n matrices. Then, det(ab) = det(a) det(b) A = ; B = A = 3-2 = 1; B = = AB = AB = AB = = -23!!! 3 14 Note: det(a+b) det(a) + det(b). Theorem: Theorem: A invertible iff det(a) 0. n n -1 1 det(a ) =. det(a) III-20

21 A = det(a) = 0 A is not invertible. Notation: When det(a) = 0, we say A is singular. When det(a) 0, we say A is nonsingular. III-21

22 [5] Inverse Some useful facts for A -1: a a (1) A = A = a nn 1/a /a /a nn B p p O p q B (2) A = A = O q p C q q (pq) (pq) 0 C 1 EX1: A = A = EX2: = A. A =? III-22

23 Definition: Let A = [a ] ; ij n n C C 1n C C 2n C = matrix of cofactors = ; and adj(a) = C t. C n1... C nn Theorem: Suppose that A is invertible. Then, -1 1 A = det(a) adj(a) Note: -1 A exists iff det(a) 0. A = C 11 = (-1) = 12; C 12 = (-1) = 6; C 13 = (-1) = C 21 = (-1) = 4; C 22 = 2; C 23 = 16; 4 0 III-23

24 C = 12; C = -10; C = C = t C = adj(a) = det(a) = = A = III-24

25 [4] Cramer s Rule Consider Ax = b where A n n. n equations and n unknowns: x 1 x = x n How can we find solutions, x 1 x n? Assume that A is invertible. Ax = b AAx = Ab I x = A b x = x 1 x n = A b n Cramer s Rule: Consider Ax = b, where A n n. Define a 11 a b 1... a 1n a 21 a b 2... a 2n x j det(a j A ) j =. Then,. det(a) a n1 a n2... b n... a nn III-25

26 x 1 + 2x 3= 6-3x 1 + 4x 2 + 6x 3 = 30 -x 1-2x 2 + 3x 3 = 8 x A = : b = 30 : x = x x A = ; A = ; A = x 1 = A1/A; x 2 = A1/A; and x 3 = A1/A. Q d = a + bp ; Q s = c + dp Then, Q = a + b P Q b P a Q = c + d P Q d P c 1 b 1 d Q P a c A x b III-26

27 a b 1 a A 1 = det(a 1) = -ad + bc; A 2= det(a 2) = c - a c d 1 c det(a) = -d + b Q det(a 1 ) det(a) adbc db adbc ; db P det(a 2 ) det(a) c a d b a c d b Y = C + I 0 + G0 C = a + by Then, Y - C = I 0+ G0 -by + C = a 1 1 Ȳ I 0 G 0 b 1 C a A x b det(a) = 1 - b I 0 G I 0 G 0 det(a 1) = = I 0+G 0+a; det(a 2) = = a+b(i 0+ G 0) a 1 b a Ȳ I 0 G 0 a C a b(i 0 G 0 ; ). 1 b 1 b Solution Outcomes for a Linear-Equation System: III-27

28 Ax = b where A. n n When b 0: If A 0, there is a unique solution x x 1 1 A = ; b = = x 2 1 If A = 0, there is an infinite number of solutions or no solution A = ; b = no solution ( x =A1/A=1/0) A = ; b = infinitely many solutions ( x 1 =A1/A=0/0) When b = 0: If A 0, there is a unique solution x = x 1 0 A = ; b = =. ( x x = 0/1) If A = 0, there is an infinite number of solutions A = ; b = infinitely many solutions. ( x = 0/0) III-28

29 [5] Leontief Input-Output Model (1) An Example of Simple Economy: 1) Three industries (industry 1, 2 and 3; say, steel, autos and food) 2) An industry's products are used for production of the industry itself and others: The products of industries 1, 2 and 3 are used for industry 1. a = $ worth of ind. 1's product required for $1 worth of ind. 1's product. 11 (a = 0.2 $.2 worth of steel is needed to produce $1 worth of steel.) 11 a = $ worth of ind. 2's product required for $1 worth of ind. 1's product 21 (a = 0.4 $.4 worth of autos is needed to produce $1 worth of steel.) 21 a = $ worth of ind. 3's product required for $1 worth of ind. 1's product 31 (a = 0.1 $.1 worth of food is needed to produce $1 worth of steel.) 31 The products of industries 1, 2 and 3 are used for industry 2. a = $ worth of ind. 1's product required for $1 worth of ind. 2's product 12 (a = 0.3 $.3 worth of steel is needed to produce $1 worth of autos.) 12 a = $ worth of ind. 2's product required for $1 worth of ind. 2's product 22 (a = 0.1 $.1 worth of autos is needed to produce $1 worth of autos.) 22 a = $ worth of ind. 3's product required for $1 worth of ind. 2's product 32 (a = 0.3 $.3 worth of food is needed to produce $1 worth of autos.) 32 The products of industries 1, 2 and 3 are used for industry 3. a = 0.2 $.2 worth of steel is needed to produce $1 worth of food. 13 a = 0.2 $.2 worth of autos is needed to produce $1 worth of food. 23 a = 0.2 $.2 worth of food is needed to produce $1 worth of food. 33 III-29

30 3) Final consumption (noninput demand): d 1 = $ worth of households' consumption of steel. d 2 = $ worth of households' consumption of autos. d 3 = $ worth of households' consumption of food. 4) The sum of input values need to produce $1 worth of industry j's (j = 1, 2, 3) products should not be greater than 1: i i i3 i1 i1 i1 a +a +a = a < 1 ; a +a +a = a < 1 ; a +a +a = a < 1. 5) Let x j be the total value of industry j's products. Then, x 1 = a11x 1 + a12x 2 + a13x 3 + d 1: a11x 1 = total value of ind. 1's products used in ind. 1. 6) Let: a x = total value of ind. 1's products used in ind x = a x + a x + a x + d ; x = a x + a x + a x + d A = , which is called "input coefficient matrix". x 1 d 1 x 2 d 2 x = ; d =. x 3 d 3 III-30

31 7) Then, Ix = Ax + d. (I-A)x = d, where I-A is called "technology matrix." If we assume (I-A) is invertible and d is given, x -1 = (I-A) d. In our example, if d = (10,5,6), we can obtain x x = x 2-1 = (I-A) d = x (2) Generalization: 1) Suppose that there are n industries. Then we can define A n n, x n 1 and d n 1. 2) Then, we have (I-A)x = b. 3) If (I-A) is invertible and b is given, we can solve for x. III-31

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