solq11 - Solutions to quiz # 11 (November 18) 1. To find...

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Unformatted text preview: Solutions to quiz # 11 (November 18) 1. To find the eigenvalues, we solve the equation 0 = det 2−λ −3 1 = (2 − λ)(6 − λ) + 3 = λ2 − 8λ + 15, 6−λ and hence √ 82 − 4 · 15 = 4 ± 1. 2 Thus there are two eigenvalues λ = 3 and λ = 5. The eigenvectors with eigenvalue 3 are non-zero solutions to the system λ= 2−3 −3 1 6−3 8± |0 −1 −→ |0 −3 1| 3| 0 1 −1 −→ 0 00 |0 , |0 1 for t = 0. 1 The eigenvectors with eigenvalue 5 are non-zero solutions to the system so the eigenvectors are t 2−5 −3 1 6−5 so the eigenvectors are t |0 −3 −→ |0 −3 1 3 1 1 |0 1 −1/3 −→ |0 0 0 for t = 0. Answer: The eigenvalues are λ = 3 with the eigenvectors t the eigenvectors t 1 3 |0 , |0 1 1 for t = 0 and λ = 5 with for t = 0. 2. We have Av = 2v and hence A2 v = A (Av) = A (2v ) = 2Av = 4v and A−1 (Av) = A−1 (2v ) =⇒ A−1 A v = 2A−1 v =⇒ A−1 v = Therefore A2 − A−1 1 v. 2 7/2 7 1 v = A2 v − A−1 v = 4v − v = v = 7 . 2 2 21/2 7/2 Answer: Bv = 7 . 21/2 1 ...
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