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hw9s - MATH 4377 ADVANCED LINEAR ALGEBRA SUMMER 2011 Key to...

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MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011, Key to HW#9 Note : All the problems below are related to Theorem 3.16 on Page 191. So please read the understand Theorem 3.16 before doing the following home- work. 1. Problem 5 Page 196. Solution . By Theorem 3.16, the third column of A is 2 1 - 1 3 - 5 0 - 1 1 = 2 3 1 . The fifth column of A is - 2 1 - 1 3 - 3 0 - 1 1 + 6 1 - 2 0 = 4 - 7 - 9 . Hence A = 1 0 2 1 4 - 1 - 1 3 - 2 - 7 3 1 1 0 - 9 . 2. Problem 7 on Page 197. Solution . 2 1 - 8 1 - 3 - 3 4 12 37 - 5 1 - 2 - 4 - 17 8 -→ 1 - 2 - 4 - 17 8 0 5 0 35 - 19 0 - 2 0 14 - 19 -→ 1 0 - 4 - 31 27 0 1 0 - 7 19 / 2 0 0 0 70 - 133 / 2 -→ 1 0 - 4 0 * 0 1 0 0 * 0 0 0 1 - 133 / 140 So { u 1 , u 2 , u 4 } is a basis for R 3 . 1

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3. Problem 9 on Page 197. Solution . 0 1 2 1 - 1 - 1 2 1 - 2 2 1 3 9 4 - 1 -→ 1 3 9 4 - 1 0 1 2 1 - 1 0 0 0 - 3 6 So { u 1 , u 2 , u 4 } is a basis for W. 4. Problem 12 on Page 197. Solution . 12(a) It is easily verified that S is a linearly independent subset of V .
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