sample_final_1_answ

sample_final_1_answ - SAMPLE FINAL 1 ANSWERS NOTE: These...

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SAMPLE FINAL 1 ANSWERS NOTE: These are only answers to the problems and not full solutions! On the final exam you will be expected to show all steps used to obtain your answer. 1. Short Answer Problems a) det A = 3 - ( - 2) = 5 so A - 1 = 1 5 ± 3 - 2 1 1 ² . b) L 1 = - 1 0 0 0 1 0 0 0 1 , L 2 = 1 0 0 0 1 0 0 0 - 1 , so L 2 L 1 = - 1 0 0 0 1 0 0 0 - 1 c) λ is an eigenvalue and ~v 6 = ~ 0 is an eigenvector of a matrix A if A~v = λ~v . d) Let ~ b be in the columnspace of A . Then there exist an ~x R n such that A~x = ~ b . But then A ~ b = A ( A~x ) = A 2 ~x = 0 ~x = ~ 0 . Hence, ~ b is in the nullspace of A . 2. a) We row-reduce A to get 1 0 0 1 0 0 b)From our row-operations in a) we have A = 1 0 0 - 5 1 0 0 0 1 1 0 0 0 1 0 2 0 1 1 0 0 0 6 0 0 0 1 1 0 0 1 0 0 . 3. Since C is square and the nullspace is a subset of R 4 , C must be 4 × 4. Since the nullspace of A has one parameter, we must have the rank of
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sample_final_1_answ - SAMPLE FINAL 1 ANSWERS NOTE: These...

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