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Unformatted text preview: 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 . b) L 1 =  1 0 0 1 0 0 1 , L 2 = 1 0 0 1 0 0 1 , so L 2 ◦ L 1 =  1 0 1 1 c) Since C is 5 × 5 the characteristic polynomial of C will be a degree 5 polynomial and hence must have at least one real root. Thus, it has at least one real eigenvalue which will give a corresponding real eigenvector. 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 = ~ . Hence, ~ b is in the nullspace of A . 2. a) We rowreduce A to get 1 0 0 1 0 0 b)From our rowoperations in a) we have A = 1 0 0 5 1 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 ....
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This note was uploaded on 11/18/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.
 Spring '08
 All
 Linear Algebra, Algebra

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