253Solutions9

# Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 2007 — Solutions for Problem Set Last 1. Problems from the textbook: a. Section 6 . 1 *26. The eigenvalues are 3 and 8 . 28. The eigenvalues are 2 , 3 , and 5 . *30. The eigenvalues are 3 , 8 , and 2 . *32. 210 *44. Easy by multilinearity: det ( kA ) = k n det ( A ) . b. Section 6 . 2 1–3. All routine. The answers are 6, 24, - 24 . *4. 9 *6. - 72 14*. 8 *16. Since the first row is twice the last, the determinant is 0 . *18. 27 *30. Just computing the determinant shows at once that it is a quadratic in t , and the top coefficient is ( b - a ) . Let’s call that quadratic f ( t ) . Making t = a or t = b results in two equal columns, so a and b must be the roots of the quadratic. Therefore, f ( t ) = ( b - a )( t - a )( t - b ) . This is really the important result. The matrix is invertible if t , a , and b are all different. 50. Here is one way to do it: subtract the second-to-last row from the last, then expand on the last row to see that det ( M n ) = det ( M n - 1 ) . Since det ( M 1 ) = 1 , we get det ( M n ) = 1 for all n . Terror is not the only condition for transcendence, though it helps. – Simon Wieseltier

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c. Section 7 . 1 1–5. All routine, I hope. *6. If Av = λv and Bv = μv , then AB ( v ) = A ( μv ) = μAv = λμv , so it’s an eigenvector with eigenvalue λμ . *8. Since we want the image of 1 0 to be 5 0 , the matrix must look like 5 b 0 d . 15–19. Easy: think geometrically. 33. Since x ( t ) = 2 t 1 1 + 6 t - 1 1 , the eigenvalues are 2 , with eigenvector 1 1 , and 6 , with eigenvector - 1 1 . So we must have a b c d 1 - 1 1 1 = 2 - 6 2 6 . Solving, we get A = 4 - 2 - 2 4 . *36. We want A 3 1 = 15 5 and A 1 2 = 10 20 . Solving, we get A = 4 3 - 2 11 . d. Section 7 . 2 1–7. Should all be routine! *8. The characteristic polynomial is - λ 2 ( λ + 3 ) , so the eigenvalues are 0 , with algebraic multiplicity 2 , and - 3 . He’s completely unspoiled by failure. – Noël Coward
*10. The characteristic polynomial is ( 1 + λ ) 2 ( 1 - λ ) , so the eigenvalues are - 1 , with algebraic multiplicity 2 , and 1 . *12. The characteristic polynomial is λ ( λ - 1 ) 2 ( λ + 1 ) , so the eigenvalues are 1 , with algebraic multiplicity 2 , 0 , and - 1 . 15. The characteristic polynomial is λ 2 - +( 1 - k ) . Using the quadratic formula, the eigenvalues are 1 ± k , so there are two different eigenvalues when k > 0 , no real eigenvalues if k < 0 , and one eigenvalue with algebraic multiplicity 2 when k = 0 .

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• Spring '07
• GHITZA
• Linear Algebra, Algebra, Eigenvalues, Eigenvalue, eigenvector and eigenspace, Orthogonal matrix, eigenvector

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