# Note that lim m r m 0 since r 1 so that lim m b m 0

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Note that lim m →∞ r m = 0 (since r < 1), so that lim m →∞ b m = 0 as well (this follows from what some calculus texts call the “squeeze theorem”). c. For a fixed i and j , let c m be the ij th entry of the matrix I n + A + A 2 + · · · + A m . By part (a), c m 1 + r + r 2 + · · · + r m < 1 1 - r . Since the c m form an increasing bounded sequence, lim m →∞ c m exists (this is a fun- damental fact of calculus). d. ( I n - A )( I n + A + A 2 + · · · + A m ) = I n + A + A 2 + · · · A m - A - A 2 - · · · - A m - A m +1 = I n - A m +1 Now let m go to infinity; use parts (b) and (c). ( I n - A )( I n + A + A 2 + · · · + A m + · · · ) = I n , so that ( I n - A ) - 1 = I n + A + A 2 + · · · + A m + · · · . 85. a. The components of the j th column of the technology matrix A give the demands industry J j makes on the other industries, per unit output of J j . The fact that the j th column sum is less than 1 means that industry J j adds value to the products it produces. b. A productive economy can satisfy any consumer demand b , since the equation ( I n - A ) x = b can be solved for the output vector x : x = ( I n - A ) - 1 b (compare with Exercise 2.3.49). 115
Chapter 2 ISM: Linear Algebra c. The output x required to satisfy a consumer demand b is x = ( I n - A ) - 1 b = ( I n + A + A 2 + · · · + A m + · · · ) b = b + Ab + A 2 b + · · · + A m b + · · · . To interpret the terms in this series, keep in mind that whatever output v the industries produce generates an interindustry demand of Av . The industries first need to satisfy the consumer demand, b . Producing the output b will generate an interindustry demand, Ab . Producing Ab in turn generates an extra interindustry demand, A ( Ab ) = A 2 b , and so forth. For a simple example, see Exercise 2.3.50; also read the discussion of “chains of in- terindustry demands” in the footnote to Exercise 2.3.49. 86. a. We write our three equations below: I = 1 3 R + 1 3 G + 1 3 B L = R - G S = - 1 2 R - 1 2 G + B , so that the matrix is P = 1 3 1 3 1 3 1 - 1 0 - 1 2 - 1 2 1 . b. R G B is transformed into R G 0 , with matrix A = 1 0 0 0 1 0 0 0 0 . c. This matrix is PA = 1 3 1 3 0 1 - 1 0 - 1 2 - 1 2 0 (we apply first A , then P .) Figure 2.45: for Problem 2.4.86d. 116
ISM: Linear Algebra Section 2.4 d. See Figure 2.45. A “diagram chase” shows that M = PAP - 1 = 2 3 0 - 2 9 0 1 0 - 1 0 1 3 . 87. a. A - 1 = 0 0 1 1 0 0 0 1 0 and B - 1 = 1 0 0 0 0 1 0 1 0 . Matrix A - 1 transforms a wife’s clan into her husband’s clan, and B - 1 transforms a child’s clan into the mother’s clan. b. B 2 transforms a women’s clan into the clan of a child of her daughter. c. AB transforms a woman’s clan into the clan of her daughter-in-law (her son’s wife), while BA transforms a man’s clan into the clan of his children. The two transformations are di ff erent. (See Figure 2.46.) Figure 2.46: for Problem 2.4.87c. d. The matrices for the four given diagrams (in the same order) are BB - 1 = I 3 , BAB - 1 = 0 0 1 1 0 0 0 1 0 , B ( BA ) - 1 = 0 1 0 0 0 1 1 0 0 , BA ( BA ) - 1 = I 3 .
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