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Elementary Linear Algebra Solutions

# Elementary Linear Algebra Solutions - SOLUTIONS TO PROBLEMS...

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SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR ALGEBRA K. R. MATTHEWS DEPARTMENT OF MATHEMATICS UNIVERSITY OF QUEENSLAND First Printing, 1991

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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. (i) 0 0 0 2 4 0 R 1 R 2 2 4 0 0 0 0 R 1 1 2 R 1 1 2 0 0 0 0 ; (ii) 0 1 3 1 2 4 R 1 R 2 1 2 4 0 1 3 R 1 R 1 - 2 R 2 1 0 - 2 0 1 3 ; (iii) 1 1 1 1 1 0 1 0 0 R 2 R 2 - R 1 R 3 R 3 - R 1 1 1 0 0 0 - 1 0 - 1 - 1 R 1 R 1 + R 3 R 3 → - R 3 R 2 R 3 1 0 0 0 1 1 0 0 - 1 R 2 R 2 + R 3 R 3 → - R 3 1 0 0 0 1 0 0 0 1 ; (iv) 2 0 0 0 0 0 - 4 0 0 R 3 R 3 + 2 R 1 R 1 1 2 R 1 1 0 0 0 0 0 0 0 0 . 3. (a) 1 1 1 2 2 3 - 1 8 1 - 1 - 1 - 8 R 2 R 2 - 2 R 1 R 3 R 3 - R 1 1 1 1 2 0 1 - 3 4 0 - 2 - 2 - 10 R 1 R 1 - R 2 R 3 R 3 + 2 R 2 1 0 4 - 2 0 1 - 3 4 0 0 - 8 - 2 R 3 - 1 8 R 3 1 0 4 2 0 1 - 3 4 0 0 1 1 4 R 1 R 1 - 4 R 3 R 2 R 2 + 3 R 3 1 0 0 - 3 0 1 0 19 4 0 0 1 1 4 . The augmented matrix has been converted to reduced row–echelon form and we read off the unique solution x = - 3 , y = 19 4 , z = 1 4 . (b) 1 1 - 1 2 10 3 - 1 7 4 1 - 5 3 - 15 - 6 9 R 2 R 2 - 3 R 1 R 3 R 3 + 5 R 1 1 1 - 1 2 10 0 - 4 10 - 2 - 29 0 8 - 20 4 59 R 3 R 3 + 2 R 2 1 1 - 1 2 10 0 - 4 10 - 2 - 29 0 0 0 0 1 . From the last matrix we see that the original system is inconsistent. 1

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(c) 3 - 1 7 0 2 - 1 4 1 2 1 - 1 1 1 6 - 4 10 3 R 1 R 3 1 - 1 1 1 2 - 1 4 1 2 3 - 1 7 0 6 - 4 10 3 R 2 R 2 - 2 R 1 R 3 R 3 - 3 R 1 R 4 R 4 - 6 R 1 1 - 1 1 1 0 1 2 - 3 2 0 2 4 - 3 0 2 4 - 3 R 1 R 1 + R 2 R 4 R 4 - R 3 R 3 R 3 - 2 R 2 1 0 3 - 1 2 0 1 2 - 3 2 0 0 0 0 0 0 0 0 . The augmented matrix has been converted to reduced row–echelon form and we read off the complete solution x = - 1 2 - 3 z, y = - 3 2 - 2 z , with z arbitrary. 4. 2 - 1 3 a 3 1 - 5 b - 5 - 5 21 c R 2 R 2 - R 1 2 - 1 3 a 1 2 - 8 b - a - 5 - 5 21 c R 1 R 2 1 2 - 8 b - a 2 - 1 3 a - 5 - 5 21 c R 2 R 2 - 2 R 1 R 3 R 3 + 5 R 1 1 2 - 8 b - a 0 - 5 19 - 2 b + 3 a 0 5 - 19 5 b - 5 a + c R 3 R 3 + R 2 R 2 - 1 5 R 2 1 2 - 8 b - a 0 1 - 19 5 2 b - 3 a 5 0 0 0 3 b - 2 a + c R 1 R 1 - 2 R 2 1 0 - 2 5 ( b + a ) 5 0 1 - 19 5 2 b - 3 a 5 0 0 0 3 b - 2 a + c . From the last matrix we see that the original system is inconsistent if 3 b - 2 a + c 6 = 0. If 3 b - 2 a + c = 0, the system is consistent and the solution is x = ( b + a ) 5 + 2 5 z, y = (2 b - 3 a ) 5 + 19 5 z, where z is arbitrary. 5. 1 1 1 t 1 t 1 + t 2 3 R 2 R 2 - tR 1 R 3 R 3 - (1 + t ) R 1 1 1 1 0 1 - t 0 0 1 - t 2 - t R 3 R 3 - R 2 1 1 1 0 1 - t 0 0 0 2 - t = B. Case 1. t 6 = 2. No solution. 2
Case 2. t = 2. B = 1 0 1 0 - 1 0 0 0 0 1 0 1 0 1 0 0 0 0 .

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Elementary Linear Algebra Solutions - SOLUTIONS TO PROBLEMS...

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