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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
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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 ·
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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