HW10-solutions - huynh(lth436 HW10 gilbert(57245 This...

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Unformatted text preview: huynh (lth436) – HW10 – gilbert – (57245) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Let A be a 2 × 2 matrix with eigenvalues 3 and 2 3 and corresponding eigenvectors v 1 = bracketleftbigg − 1 − 2 bracketrightbigg , v 2 = bracketleftbigg − 1 bracketrightbigg . Let { x k } be a solution of the difference equa- tion x k +1 = A x k , x = bracketleftbigg 10 2 bracketrightbigg . Compute x 1 . 1. x 1 = bracketleftbigg 9 6 bracketrightbigg correct 2. x 1 = bracketleftbigg − 6 9 bracketrightbigg 3. x 1 = bracketleftbigg 9 − 6 bracketrightbigg 4. x 1 = bracketleftbigg − 6 − 9 bracketrightbigg 5. x 1 = bracketleftbigg 6 9 bracketrightbigg 6. x 1 = bracketleftbigg − 9 − 6 bracketrightbigg Explanation: To find x 1 we must compute A x . Now, ex- press x in terms of v 1 and v 2 . That is, find c 1 and c 2 such that x = c 1 v 1 + c 2 v 2 . This is certainly possible because the eigenvectors v 1 and v 2 are linearly independent (by in- spection and also because they correspond to distinct eigenvalues) and hence form a basis for R 2 . The row reduction [ v 1 v 2 x ] = bracketleftbigg − 1 − 1 10 − 2 2 bracketrightbigg ∼ bracketleftbigg 1 − 1 1 − 9 bracketrightbigg shows that x = − v 1 − 9 v 2 . Since v 1 and v 2 are eigenvectors (for the eigenvalues 3 and 2 3 respectively): x 1 = A x = A ( − v 1 − 9 v 2 ) = − A v 1 − 9 A v 2 = − 3 v 1 − 9 · 2 3 v 2 = bracketleftbigg 3 6 bracketrightbigg + bracketleftbigg 6 bracketrightbigg = bracketleftbigg 9 6 bracketrightbigg . Consequently, x 1 = bracketleftbigg 9 6 bracketrightbigg . 002 10.0 points Let A be a 2 × 2 matrix with eigenvalues 4 and 3 and corresponding eigenvectors v 1 = bracketleftbigg 1 1 bracketrightbigg , v 2 = bracketleftbigg 2 bracketrightbigg . Determine the solution { x k } of the difference equation x k +1 = A x k , x = bracketleftbigg 2 − 4 bracketrightbigg . 1. x k = − 8 (4) k v 1 − 3 (3) k v 2 2. x k = − 4 (4) k v 1 + 6 (3) k v 2 3. x k = − 8 (4) k v 1 + 3 (3) k v 2 4. x k = − 4 (4) k v 1 + 3 (3) k v 2 correct 5. x k = − 8 (4) k v 1 + 6 (3) k v 2 6. x k = − 4 (4) k v 1 − 6 (3) k 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 x = c 1 v 1 + c 2 v 2 huynh (lth436) – HW10 – gilbert – (57245) 2 To compute c 1 and c 2 we apply row reduction to the augmented matrix [ v 1 v 2 x ] = bracketleftbigg 1 2 2 1 − 4 bracketrightbigg ∼ bracketleftbigg 1 − 4 1 3 bracketrightbigg . This shows that c 1 = − 4, c 2 = 3 and x = − 4 v 1 + 3 v 2 . Since v 1 and v 2 are eigen- vectors corresponding to the eigenvalues 4 and 3 respectively, x k = A k x = A k ( − 4 v 1 + 3 v 2 ) = − 4 A k v 1 + 3 A k v 2 = − 4 (4) k v 1 + 3 (3) k v 2 for k ≥ 0. It follows that A x k = − 4 (4) k A v 1 + 3 (3) k A v 2 = − 4 (4) k +1 v 1 + 3 (3)...
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