HW11-solutions

# HW11-solutions - huang(dh34953 HW11 gilbert(54160 This...

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huang (dh34953) – HW11 – gilbert – (54160) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Let A be a 2 × 2 matrix with eigenvalues 2 and 1 and corresponding eigenvectors v 1 = b 1 3 B , v 2 = b 1 2 B . Let { x k } be a solution oF the di±erence equa- tion x k +1 = A x k , x 0 = b 1 1 B . Compute x 1 . 1. x 1 = b 4 10 B 2. x 1 = b 4 10 B 3. x 1 = b 10 4 B 4. x 1 = b 4 10 B correct 5. x 1 = b 10 4 B 6. x 1 = b 10 4 B Explanation: To fnd x 1 we must compute A x 0 . Now, ex- press x 0 in terms oF v 1 and v 2 . That is, fnd 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 ] = b 1 1 1 3 2 1 B b 1 0 1 0 1 2 B shows that x 0 = v 1 + 2 v 2 . Since v 1 and v 2 are eigenvectors (For the eigenvalues 2 and 1 respectively): x 1 = A x 0 = A ( v 1 + 2 v 2 ) = A v 1 + 2 A v 2 = 2 v 1 + 2 · − 1 v 2 = b 2 6 B + b 2 4 B = b 4 10 B . Consequently, x 1 = b 4 10 B . 002 10.0 points Let A be a 3 × 3 matrix with eigenvalues 2 , 1 , and 1 and corresponding eigenvectors v 1 = 2 4 6 , v 2 = 1 1 9 , v 3 = 2 7 3 . IF { x k } is the solution oF the di±erence equa- tion x k +1 = A x k , x 0 = 5 16 0 , determine x 1 . 1. x 1 = 18 10 5 2. x 1 = 18 10 5 3. x 1 = 5 10 18 4. x 1 = 18 10 5 5. x 1 = 5 10 18 correct 6. x 1 = 5 10 18

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huang (dh34953) – HW11 – gilbert – (54160) 2 Explanation: To fnd x 1 we must compute A x 0 . First, we express express x 0 in terms o± 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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