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# ism1 - Chapter 1 Linear Equations and Matrices Section 1.1...

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Chapter 1 Linear Equations and Matrices Section 1.1, p. 8 2. x = 1, y = 2, z = 2. 4. No solution. 6. x = 13 + 10 z , y = 8 8 z , z = any real number . 8. No solution. 10. x = 2, y = 1. 12. No solution. 14. x = 1, y = 2, z = 2. 16. (c) Yes. (d) Yes. 18. x = 2, y = 1, z = 0. 20. There is no such value of r . 22. Zero, infinitely many, zero. 24. 1.5 tons of regular and 2.5 tons of special plastic. 26. 20 tons of 2-minute developer and a total of 40 tons of 6-minute and 9-minute developer. 28. \$7000, \$14,000, \$3000. T.1. The same numbers s j satisfy the system when the p th equation is written in place of the q th equation and vice versa. T.2. If s 1 , s 2 , . . . , s n is a solution to (2), then the i th equation of (2) is satisfied: a i 1 s 1 + a i 2 s 2 + · · · + a in s n = b i . Then for any r = 0, ra i 1 s 1 + ra i 2 s 2 + · · · + ra in s n = rb i . Hence s 1 , s 2 , . . . , s n is a solution to the new system. Conversely, for any solution s 1 , s 2 , . . . , s n to the new system, ra i 1 s 1 + · · · + ra in s n = rb i , and dividing both sides by nonzero r we see that s 1 , . . . , s n must be a solution to the original linear system. T.3. If s 1 , s 2 , . . . , s n is a solution to (2), then the p th and q th equations are satisfied: a p 1 s 1 + · · · a pn = b p a q 1 s 1 + · · · a qn = b q .

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2 Chapter 1 Thus, for any real number r , ( a p 1 + ra q 1 ) s 1 + · · · + ( a pn + ra qn ) s n = b p + rb q and so s 1 , . . . , s n is a solution to the new system. Conversely, any solution to the new system is also a solution to the original system (2). T.4. Yes; x = 0, y = 0 is a solution for any values of a , b , c , and d . Section 1.2, p. 19 2. a = 3, b = 1, c = 8, d = 2. 4. (a) C + E = E + C = 5 5 8 4 2 9 5 3 4 . (b) Impossible. (c) 7 7 0 1 . (d) 9 3 9 12 3 15 6 3 9 . (e) 0 10 9 8 1 2 5 4 3 . (f) Impossible. 6. (a) A T = 1 2 2 1 3 4 , ( A T ) T = 1 2 3 2 1 4 . (b) 5 4 5 5 2 3 8 9 4 . (c) 6 10 11 17 . (d) 0 4 4 0 . (e) 3 4 6 3 9 10 . (f) 17 2 16 6 . 8. Yes: 2 1 0 0 1 + 1 1 0 0 0 = 3 0 0 2 . 10. λ 1 2 3 6 λ + 2 3 5 2 λ 4 . 12. (a) 0 0 1 1 . (b) 1 1 1 0 . (c) 1 1 0 1 . (d) 0 1 0 1 . (e) 1 1 0 1 . 14. v = 0 0 0 0 . T.1. Let A and B each be diagonal n × n matrices. Let C = A + B , c ij = a ij + b ij . For i = j , a ij and b ij are each 0, so c ij = 0. Thus C is diagonal. If D = A B , d ij = a ij b ij , then d ij = 0. Therefore D is diagonal. T.2. Following the notation in the solution of T.1 above, let A and B be scalar matrices, so that a ij = 0 and b ij = 0 for i = j , and a ii = a , b ii = b . If C = A + B and D = A B , then by Exercise T.1, C and D are diagonal matrices. Moreover, c ii = a ii + b ii = a + b and d ii = a ii b ii = a b , so C and D are scalar matrices. T.3. (a) 0 b c c e c b 0 0 e c 0 0 . (b) 2 a c + b e + c b + c 2 d 2 e c + e 2 e 2 f . (c) Same as (b). T.4. Let A = a ij and C = c ij , so c ij = ka ij . If ka ij = 0, then either k = 0 or a ij = 0 for all i , j .
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