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ANSWERS AND SUGGESTIONS FOR SELECTED EXERCISES Chapter 1 Section 1.1, p.4 1. (a) no (b) no (c) no (d) no (e) yes. 3. (a) 1 ; 1 g (b) ( 1 ; 1 ; 1 ± i p 3 2 ; 1 ± i p 3 2 ) (c) Z 7 : 5. x = 1 ; y = 0 and z = 2 : Section 1.2, p.9 1. (a) 2 4 1 4 4 7 9 12 3 5 (b) " 0 1 3 # (c) 2 6 6 4 2 0 3 4 4 5 5 6 3 7 7 5 (d) 2 4 0 1 1 0 1 1 3 5 : 3. a = 2 : Section 1.3, p.25 (b) 0 6 0 2 6 1 ± (c) 1 12 5 7 ± 3. x = y = 2 and z = 1. 5. The only triple product is BCA = 6 3 3 0 0 0 ± : 7. (a) false; for example take A = 1 0 0 1 ± : (b) false; since AB 6 = BA for some matrices. (c) true (d) false; for example take A = 1 0 0 0 ± and B = & 0 0 1 1 ± : 9. A 6 = 2 4 0 0 0 0 0 0 0 0 0 3 5 : 1

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11. All solutions are of the form & 0 a 1 a 0 ± where a 2 F &f 0 g : Thus the number of solutions is equal to the number of non-zero elements in F: 13. a 0 0 a ± : 15. Rows of BA are R 1 3 R 2 + R 3 ; 2 R 2 + 2 R 3 ; R 1 + R 2 . Section 1.4, p.36 1. diag (1 ; 1 ; 9 ; 16) ; diag (2 ; 2 ; 9 ; 0) ; diag (3 ; 3 ; 0 ; 4) ; diag ( 3 ; 3 ; 0 ; 16). 3. There are 81 matrices satisfying the given equality, and 3 of them are scalar matrices. 7. x = 1 ; y = 0 ; z = 2 : 13. 1 0 0 1 ± ; 1 0 0 0 ± ; 1 1 0 0 ± ; 1 2 0 0 ± ; 0 1 0 0 ± ; 0 0 0 0 ± : Section 1.5, p.56 1. One of them is E = 2 6 6 4 0 1 1 1 2 0 0 1 3 5 0 0 0 0 0 0 0 0 0 0 3 7 7 5 . 3. (a) 2 6 6 4 1 0 0 0 2 1 0 0 0 0 1 0 0 0 0 1 3 7 7 5 (b) 2 6 6 4 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 3 7 7 5 (c) 2 6 6 4 1 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 3 7 7 5 : (d) 2 6 6 4 1 0 0 0 0 1 0 0 0 0 2 0 0 0 0 1 3 7 7 5 (e) 2 6 6 4 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 3 7 7 5 (f) 2 6 6 4 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 3 7 7 5 : 5. We may take E 1 = 2 4 1 0 0 1 1 0
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