2 k v 3 for k 0 it follows that a x k 3 3 k a v 1 3 2

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Unformatted text preview: 2) k v 3 for k ≥ 0. It follows that A x k = − 3 (3) k A v 1 − 3 (2) k A v 2 + 2 ( − 2) k A v 3 = − 3 (3) k +1 v 1 − 3 (2) k +1 v 2 + 2 ( − 2) k +1 v 3 = x k +1 . This shows that { x k } solves the difference equation. Consequently, x k = − 3 (3) k v 1 − 3 (2) k v 2 + 2 ( − 2) k v 3 . 005 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3, 2 5 , and 1 3 and corresponding eigenvectors v 1 = 1 − 3 − 2 , v 2 = 2 − 5 − 5 , v 3 = 1 − 2 . Determine the solution { x k } of the difference equation x k +1 = A x k , x = − 4 8 6 . 1. x k = − 2 (3) k v 1 − 2 ( 2 5 ) k v 2 − 2 ( 1 3 ) k v 3 2. x k = 2 (3) k v 1 + 2 ( 2 5 ) k v 2 + 2 ( 1 3 ) k v 3 3. x k = 2 (3) k v 1 + 2 ( 2 5 ) k v 2 − 2 ( 1 3 ) k v 3 4. x k = 2 (3) k v 1 − 2 ( 2 5 ) k v 2 − 2 ( 1 3 ) k v 3 correct 5. x k = − 2 (3) k v 1 − 2 ( 2 5 ) k v 2 + 2 ( 1 3 ) k v 3 6. x k = − 2 (3) k v 1 + 2 ( 2 5 ) k v 2 + 2 ( 1 3 ) k v 3 Explanation: huynh (lth436) – HW10 – gilbert – (57245) 4 Since v 1 , v 2 , and v 3 are eigenvectors cor- responding to distinct eigenvalues of A , they form an eigenbasis for R 3 . Thus x = c 1 v 1 + c 2 v 2 + c 3 v 3 To compute c 1 , c 2 , and c 3 we apply row re- duction to the augmented matrix [ v 1 v 2 v 3 x ] = 1 2 1 − 4 − 3 − 5 − 2 8 − 2 − 5 6 ∼ 1 2 1 − 2 1 − 2 . This shows that c 1 = 2, c 2 = − 2, c 3 = − 2 and x = 2 v 1 − 2 v 2 − 2 v 3 . Since v 1 , v 2 , and v 3 are eigenvectors corresponding to the eigenvalues 3, 2 5 , 1 3 respectively, x k = A k x = A k (2 v 1 − 2 v 2 − 2 v 3 ) = 2 A k v 1 − 2 A k v 2 − 2 A k v 3 = 2 (3) k v 1 − 2 ( 2 5 ) k v 2 − 2 ( 1 3 ) k v 3 for k ≥ 0. It follows that A x k = 2 (3) k A v 1 − 2 ( 2 5 ) k A v 2 − 2 ( 1 3 ) k A v 3 = 2 (3) k +1 v 1 − 2 ( 2 5 ) k +1 v 2 − 2 ( 1 3 ) k +1 v 3 = x k +1 . This shows that { x k } solves the difference equation. Consequently, x k = 2 (3) k v 1 − 2 ( 2 5 ) k v 2 − 2 ( 1 3 ) k v 3 . 006 10.0 points Find the solution of the differential equa- tion d u dt = A u ( t ) , u (0) = bracketleftbigg 2 − 3 bracketrightbigg when A is a 2 × 2 matrix with eigenvalues 5 and 4 and corresponding eigenvectors v 1 = bracketleftbigg − 1 1 bracketrightbigg , v 2 = bracketleftbigg 1 bracketrightbigg . 1. u ( t ) = − 6 e 5 t v 1 − 2 e 4 t v 2 2. u ( t ) = − 3 e 5 t v 1 − 2 e 4 t v 2 3. u ( t ) = − 6 e 5 t v 1 + e 4 t v 2 4. u ( t ) = − 3 e 5 t v 1 + 2 e 4 t v 2 5. u ( t ) = − 3 e 5 t v 1 − e 4 t v 2 correct 6. u ( t ) = − 6 e 5 t v 1 − e 4 t v 2 Explanation: Since v 1 and v 2 are eigenvectors corre- sponding to distinct eigenvalues of A , they form an eigenbasis for R 2 . Thus u (0) = c 1 v 1 + c 2 v 2 To compute c 1 and c 2 we apply row reduction to the augmented matrix [ v 1 v 2 u (0) ] = bracketleftbigg − 1 1 2 1 − 3 bracketrightbigg ∼ bracketleftbigg 1 − 3 1 − 1 bracketrightbigg ....
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