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assign8_practice_soln

# assign8_practice_soln - Math 136 Practice Problems 8 1 For...

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Math 136 Practice Problems # 8 1. For each of the following matrices, find the inverse, or show that the matrix is not invertible. a) A = 1 2 1 0 - 2 4 3 4 4 Solution: To determine if A is invertible we write [ A | I ] and row reduce: 1 2 1 1 0 0 0 - 2 4 0 1 0 3 4 4 0 0 1 1 0 0 - 4 - 2 / 3 5 / 3 0 1 0 2 1 / 6 - 2 / 3 0 0 1 1 1 / 3 - 1 / 3 Hence, A - 1 = - 4 - 2 / 3 5 / 3 2 1 / 6 - 2 / 3 1 1 / 3 - 1 / 3 . b) B = 4 - 1 - 1 0 2 3 - 3 2 3 Solution: To determine if B is invertible we write [ B | I ] and row reduce: 4 - 1 - 1 1 0 0 0 2 3 0 1 0 - 3 2 3 0 0 1 1 0 0 0 1 / 3 - 1 / 3 0 1 0 - 3 3 - 4 0 0 1 2 - 5 / 3 8 / 3 Hence, B - 1 = 0 1 / 3 - 1 / 3 - 3 3 - 4 2 - 5 / 3 8 / 3 . c) C = - 1 0 - 1 2 1 - 1 - 2 4 - 3 - 1 5 2 - 1 - 2 4 4 Solution: To determine if C is invertible we write [ C | I ] and row reduce: - 1 0 - 1 2 1 0 0 0 1 - 1 - 2 4 0 1 0 0 - 3 - 1 5 2 0 0 1 0 - 1 - 2 4 4 0 0 1 - 1 0 - 1 2 1 0 0 0 0 - 1 - 3 6 1 1 0 0 0 0 11 - 10 - 4 - 1 1 0 0 0 0 0 1 - 1 - 1 1 Since the RREF of C will not be I , it follows that C is not invertible. 1

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2 2. Let B = 1 - 2 2 - 1 1 0 3 - 2 - 3 . Find B - 1 and use it to solve B~x = ~ d , where ~ d = 1 - 2 3 . Solution: To find the inverse of B we write [ B | I ] and row reduce: 1 - 2 2 1 0 0 - 1 1 0 0 1 0 3 - 2 - 3 0 0 1 1 0 0 - 3 - 10 - 2 0 1 0 - 3 - 9 - 2 0 0 1 - 1 - 4 - 1 Therefore B - 1 = - 3 - 10 - 2 - 3 - 9 - 2 - 1 - 4 - 1 . We have ~x = B - 1 ( B~x ) = B - 1 1 - 2 3 = - 3 - 10 - 2 - 3 - 9 - 2 - 1 - 4 - 1 1 - 2 3 = 11 9 4 .
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