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253Solutions9

# Linear Algebra with Applications (3rd Edition)

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ma253 ,Fa l l 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 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 frst 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 , andthetopcoe Ffc ientis ( b - a ) . Let’s call that quadratic f ( t ) .Mak ing 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. Thematr ixisinvert ib leiF t , a 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 ) .S in ce 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 ,soi t ’san eigenvector with eigenvalue λμ . *8. S incewewanttheimageof ± 1 0 ² to be ± 5 0 ² , the matrix must look like ± 5b 0d ² . 15–19. Easy: think geometrically. 33. Since x ( t 2 t ± 1 1 ² + 6 t ± - 1 1 ² , the eigenvalues are 2 , with eigenvector ± 1 1 ² ,and 6 , with eigenvector ± - 1 1 ² .Sowe must have ± ab cd ²± 1 - 1 11 ² = ± 2 - 6 26 ² . Solving, we get A = ± 4 - 2 - 24 ² . *36. We want A ± 3 1 ² = ± 15 5 ² and A ± 1 2 ² = ± 10 20 ² .So lv ing ,weget A = ± 43 - 21 1 ² . d. Section 7 . 2 1–7. Should all be routine! *8. The characteristic polynomial is - λ 2 ( λ + 3 ) , so the eigenvalues are 0 ,w i th algebraic multiplicity 2 - 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 - 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

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253Solutions9 - ma253 Fall 2007 Solutions for Problem Set...

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