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SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ALGEBRA K. R. MATTHEWS DEPARTMENT OF MATHEMATICS UNIVERSITY OF QUEENSLAND First Printing, 1991 CONTENTS PROBLEMS 1.6 ............................................ 1 PROBLEMS 2.4 ............................................ 12 PROBLEMS 2.7 ............................................ 18 PROBLEMS 3.6 ............................................ 32 PROBLEMS 4.1 ............................................ 45 PROBLEMS 5.8 ............................................ 58 PROBLEMS 6.3 ............................................ 69 PROBLEMS 7.3 ............................................ 83 PROBLEMS 8.8 ............................................ 91 i SECTION 1.6 ¸ ¸ · · 2 4 0 0 0 0 1 2 0 1 R1 → 2 R1 R1 ↔ R 2 ; 2. (i) 0 0 0 2 4 0 0 0 0 · ¸ · ¸ · ¸ 0 1 3 1 2 4 1 0 −2 (ii) R1 ↔ R 2 R1 → R1 − 2R2 ; 1 2 4 0 1 3 0 1 3 1 1 0 1 1 1 R → R2 − R1 0 0 −1 (iii) 1 1 0 2 R3 → R 3 − R 1 0 −1 −1 1 0 0 1 0 0 1 0 0 R1 → R 1 + R 3 R → R2 + R3 1 2 0 1 0 ; R3 → −R3 0 1 R3 → −R3 R2 ↔ R 3 0 0 −1 0 0 1 2 0 0 1 0 0 R → R3 + 2R1 0 0 0 . (iv) 0 0 0 3 R1 → 12 R1 −4 0 0 0 0 0 1 1 1 2 1 1 1 2 R → R2 − 2R1 0 1 −3 4 3 −1 8 2 3. (a) 2 R3 → R 3 − R 1 0 −2 −2 −10 1 −1 −1 −8 1 0 4 −2 1 0 4 2 R1 → R 1 − R 2 0 1 −3 4 R3 → −1 R3 0 1 −3 4 8 R3 → R3 + 2R2 0 0 −8 −2 0 0 1 14 1 0 0 −3 R1 → R1 − 4R3 . 0 1 0 19 4 R2 → R2 + 3R3 1 0 0 1 4 · ¸ The augmented matrix has been converted to reduced row–echelon form and we read off the unique solution x = −3, y = 19 , z = 14 . 4 1 1 −1 2 10 1 1 −1 2 10 R → R2 − 3R1 7 4 1 2 0 −4 10 −2 −29 (b) 3 −1 R3 → R3 + 5R1 −5 3 −15 −6 9 0 8 −20 4 59 1 1 −1 2 10 R3 → R3 + 2R2 0 −4 10 −2 −29 . 0 0 0 0 1 From the last matrix we see that the original system is inconsistent. 1