HW11-solutions - farias(df8272 HW11 gilbert(56780 This...

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farias (df8272) – HW11 – gilbert – (56780) 1 This print-out should have 20 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 1 and 1 and corresponding eigenvectors v 1 = 4 12 , v 2 = 1 2 . Let { x k } be a solution of the di ff erence equa- tion x k +1 = A x k , x 0 = 5 14 . Compute x 1 . 1. x 1 = 3 10 2. x 1 = 10 3 3. x 1 = 10 3 4. x 1 = 3 10 correct 5. x 1 = 3 10 6. x 1 = 10 3 Explanation: To find x 1 we must compute A x 0 . Now, ex- press x 0 in terms of v 1 and v 2 . That is, find c 1 and c 2 such that x 0 = 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 0 ] = 4 1 5 12 2 14 1 0 1 0 1 1 shows that x 0 = v 1 + v 2 . Since v 1 and v 2 are eigenvectors (for the eigenvalues 1 and 1 respectively): x 1 = A x 0 = A ( v 1 + v 2 ) = A v 1 + A v 2 = 1 v 1 + 1 v 2 = 4 12 + 1 2 = 3 10 . Consequently, x 1 = 3 10 . 002 10.0 points Let A be a 2 × 2 matrix with eigenvalues 5 and 2 and corresponding eigenvectors v 1 = 2 6 , v 2 = 2 9 . Determine the solution { x k } of the di ff erence equation x k +1 = A x k , x 0 = 2 12 . 1. x k = 2 (5) k v 1 2 (2) k v 2 2. x k = (5) k v 1 2 (2) k v 2 correct 3. x k = 2 (5) k v 1 + 2 (2) k v 2 4. x k = (5) k v 1 + 4 (2) k v 2 5. x k = (5) k v 1 4 (2) k v 2 6. x k = 2 (5) k v 1 4 (2) 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 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 x 0 ] = 2 2 2 6 9 12 1 0 1 0 1 2 .
farias (df8272) – HW11 – gilbert – (56780) 2 This shows that c 1 = 1, c 2 = 2 and x 0 = v 1 2 v 2 . Since v 1 and v 2 are eigenvec- tors corresponding to the eigenvalues 5 and 2 respectively, x k = A k x 0 = A k ( v 1 2 v 2 ) = A k v 1 2 A k v 2 = (5) k v 1 2 (2) k v 2 for k 0. It follows that A x k = (5) k A v 1 2 (2) k A v 2 = (5) k +1 v 1 2 (2) k +1 v 2 = x k +1 . This shows that { x k } solves the di ff erence equation. Consequently, x k = (5) k v 1 2 (2) k v 2 . 003 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3 , 2 , and 3 and corresponding eigenvectors v 1 = 2 4 4 , v 2 = 2 7 1 , v 3 = 1 4 3 . If { x k } is the solution of the di ff erence equa- tion x k +1 = A x k , x 0 = 4 18 8 , determine x 1 . 1. x 1 = 4 8 10 correct 2. x 1 = 4 8 10 3. x 1 = 10 8 4 4. x 1 = 4 8 10 5. x 1 = 10 8 4 6. x 1 = 10 8 4 Explanation: To find x 1 we must compute A x 0 . First, we express express x 0 in terms of v 1 , v 2 , and v 3 : x 0 = c 1 v 1 + c 2 v 2 + c 3 v 3 .

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