# Ism1 - Chapter 1 Linear Equations and Matrices Section 1.1 p 8 2 x = 1 y = 2 z =-2 4 No solution 6 x = 13 10z y =-8 8z z = any real number 8 No

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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 y = 1. 12. No solution. 14. x = 1, y z = 2. 16. (c) Yes. (d) Yes. 18. x y 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 + i 2 s 2 + + in s n = rb i . Hence s 1 2 n is a solution to the new system. Conversely, for any solution s ± 1 ± 2 ± n to the new system, i 1 s ± 1 + + in s ± n = i , and dividing both sides by nonzero r we see that s ± 1 ± n must be a solution to the original linear system. T.3. If s 1 2 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 + 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 58 429 534 . (b) Impossible. (c) ± 7 7 01 ² . (d) 93 9 12 3 15 6 3 9 . (e) 0 9 8 1 2 5 43 . (f) Impossible. 6. (a) A T = 12 21 34 ,( A T ) T = ± 123 214 ² . (b) 545 523 894 . (c) ± 61 0 11 17 ² . (d) ± 0 4 40 ² . (e) 63 91 0 .( f ) ± 17 2 16 6 ² . 8. Yes: 2 ± 10 ² +1 ± 00 ² = ± 30 02 ² . 10. λ 1 2 3 6 λ +2 3 5 2 λ 4 . 12. (a) ± 11 ² . (b) ± ² . (c) ± ² . (d) ± ² . (e) ± ² . 14. v = ³ 0000 ´ . T.1. Let A and B each be diagonal n × n matrices. Let C = A + B , c ij = a ij + b ij .F o r 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 and b ij = 0 for i ± = j , and a ii = a , b ii = b .I f 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 cc e c b e c . (b) 2 ac + be + c b + c 2 d 2 e c + e 2 e 2 f . (c) Same as (b). T.4. Let A = ³ a ij ´ and C = ³ c ij ´ c ij = ka ij f ij = 0, then either k =0or a ij = 0 for all i , j . T.5. (a) Let A = ³ a ij ´ and B = ³ b ij ´ be upper triangular matrices, and let C = A + B . Then for i>j , c ij = a ij + b ij = 0 + 0 = 0, and thus C is upper triangular. Similarly, if D = A B , then for , d ij = a ij b ij 0 = 0, so D is upper triangular.
Section 1.2 3 (b) Proof is similar to that for (a).

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## This note was uploaded on 06/25/2008 for the course MATH 22a taught by Professor Chuchel during the Spring '08 term at UC Davis.

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Ism1 - Chapter 1 Linear Equations and Matrices Section 1.1 p 8 2 x = 1 y = 2 z =-2 4 No solution 6 x = 13 10z y =-8 8z z = any real number 8 No

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