129a_ans_5.1 - E A (3) is -1 1 1 5.1 #18 the eigenvalues...

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Spring 2010 Answer of Section 5.1 5.1 #2 YES, because det ±" 7 3 3 - 1 # - ( - 2) I 2 ! = 0. 5.1 #5 YES, because 3 7 9 - 4 - 5 1 2 4 43 4 - 3 1 = 0 0 0 and the eigenvalue is 0. 5.1 #13 a basis of E A (1) is 0 1 0 a basis of E A (2) is - 1 2 2 a basis of
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Unformatted text preview: E A (3) is -1 1 1 5.1 #18 the eigenvalues are 4, 0, -3 5.1 #29 Let e be the vector with all entries equal 1. Then A e = s e . Hence e is an eigenvector with eigenvalue s ....
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