Note that limm→∞rm= 0 (sincer <1), so that limm→∞bm= 0 as well (this followsfrom what some calculus texts call the “squeeze theorem”).c. For a fixediandj, letcmbe theijth entry of the matrixIn+A+A2+· · ·+Am. Bypart (a),cm≤1 +r+r2+· · ·+rm<11-r.Since thecmform an increasing bounded sequence, limm→∞cmexists (this is a fun-damental fact of calculus).d. (In-A)(In+A+A2+· · ·+Am) =In+A+A2+· · ·Am-A-A2-· · ·-Am-Am+1=In-Am+1Now letmgo to infinity; use parts (b) and (c). (In-A)(In+A+A2+· · ·+Am+· · ·) =In,so that(In-A)-1=In+A+A2+· · ·+Am+· · ·.85. a. The components of thejth column of the technology matrixAgive the demandsindustryJjmakes on the other industries, per unit output ofJj. The fact that thejth column sum is less than 1 means that industryJjadds valueto the products itproduces.b. A productive economy can satisfy any consumer demandb, since the equation(In-A)x=bcan be solved for the output vectorx:x= (In-A)-1b(compare withExercise 2.3.49).115
Chapter 2ISM:Linear Algebrac. The outputxrequired to satisfy a consumer demandbisx= (In-A)-1b= (In+A+A2+· · ·+Am+· · ·)b=b+Ab+A2b+· · ·+Amb+· · ·.To interpret the terms in this series, keep in mind that whatever outputvthe industriesproduce generates an interindustry demand ofAv.The industries first need to satisfy the consumer demand,b. Producing the outputbwill generate an interindustry demand,Ab. ProducingAbin turn generates an extrainterindustry demand,A(Ab) =A2b, 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 18.104.22.168. a. We write our three equations below:I=13R+13G+13BL=R-GS=-12R-12G+B, so that the matrix isP=1313131-10-12-121.b.RGBis transformed intoRG0, with matrixA=100010000.c. This matrix isPA=131301-10-12-120(we apply firstA, thenP.)Figure 2.45: for Problem 2.4.86d.116
ISM:Linear AlgebraSection 2.4d. See Figure 2.45. A “diagram chase” shows thatM=PAP-1=230-29010-1013.87. a.A-1=001100010andB-1=100001010.MatrixA-1transforms a wife’s clan into her husband’s clan, andB-1transforms achild’s clan into the mother’s clan.b.B2transforms a women’s clan into the clan of a child of her daughter.c.ABtransforms 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 transformationsare different. (See Figure 2.46.)Figure 2.46: for Problem 2.4.87c.d. The matrices for the four given diagrams (in the same order) areBB-1=I3,BAB-1=001100010, B(BA)-1=010001100, BA(BA)-1=I3.