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chapter4solutions

chapter4solutions - Math 152 Chapter 4 solutions Problem...

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Math 152 Chapter 4 solutions 1 Problem 4.1: Show that 1 1 and 1 - 1 are eigenvectors for the matrix 1 1 1 1 . What are the corresponding eigenvalues? Answer: We compute 1 1 1 1 1 1 = 2 2 = 2 1 1 , so 1 1 is an eigenvector with eigenvalue 2. Also 1 1 1 1 1 - 1 = 0 0 = 0 1 - 1 , so 1 - 1 is an eigenvector with eigenvalue 0. Problem 4.2: Suppose P is a projection matrix. What are the eigenvalues and eigenvectors of P ? Answer: If P projects onto some line, then a vector x lying on that line doesn’t get changed by P so P x = x and x is an eigenvector with eigenvalue 1. On the other hand, if x is perpendicular to the line, then P x = 0 = 0 x so x is an eigenvector with eigenvalue 0. Problem 4.3: Find the eigenvalues and eigenvectors for a ) 0 3 3 0 b ) - 2 - 8 4 10 c ) 29 - 10 105 - 36 d ) - 9 - 14 7 12 Answer: a) det( A - λI ) = λ 2 - 9 = ( λ - 3)( λ + 3) so the eigenvalues are λ 1 = 3 and λ 2 = - 3. To find the eigenvector for λ 1 = 3 we must solve the homegeneous equation with matrix - 3 3 3 - 3 . The matrix reduces to - 3 3 0 0 and the eigenvector is x 1 = 1 1 . To find the eigenvector for λ 2 = - 3 we must solve the homegeneous equation with matrix 3 3 3 3 . The matrix reduces to 3 3 0 0 and the eigenvector is x 2 = 1 - 1 . b) det( A - λI ) = λ 2 - 8 λ + 12 = ( λ - 2)( λ - 6) so the eigenvalues are λ 1 = 3 and λ 2 = - 3. To find the eigenvector for λ 1 = 2 we must solve the homegeneous equation with matrix - 4 - 8 4 8 . The matrix reduces to - 4 - 8 0 0 and the eigenvector is x 1 = 2 - 1 . To find the eigenvector for λ 2 = 6 we must solve the homegeneous equation with matrix - 8 - 8 4 4 . The matrix reduces to - 8 - 8 0 0 and the eigenvector is x 2 = 1 - 1 . c) det( A - λI ) = λ 2 + 7 λ + 6 = ( λ + 6)( λ + 1) so the eigenvalues are λ 1 = - 6 and λ 2 = - 1. To find the eigenvector for λ 1 = - 6 we must solve the homegeneous equation with matrix 35 - 10 105 - 30 . The matrix reduces to 35 - 10 0 0 and the eigenvector is x 1 = 2 7 . To find the eigenvector for λ 2 = - 1 we must solve the homegeneous equation with matrix 30 - 10 105 - 35 . The matrix reduces to 30 - 10 0 0 and the eigenvector is x 2 = 1 3 .
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2 Math 152 Chapter 4 solutions d) det( A - λI ) = λ 2 - 3 λ - 10 = ( λ + 2)( λ - 5) so the eigenvalues are λ 1 = - 2 and λ 2 = 5. To find the eigenvector for λ 1 = - 2 we must solve the homegeneous equation with matrix - 7 - 14 7 14 . The matrix reduces to - 7 - 14 0 0 and the eigenvector is x 1 = 2 - 1 . To find the eigenvector for λ 2 = 5 we must solve the homegeneous equation with matrix - 14 - 14 7 7 . The matrix reduces to - 14 - 14 0 0 and the eigenvector is x 2 = 1 - 1 . Problem 4.4: Find the eigenvalues and eigenvectors for a ) 0 - 1 1 1 0 2 2 0 2 b ) 1 1 1 1 0 - 2 1 - 1 1 c ) 7 - 9 - 15 0 4 0 3 - 9 - 11 d ) 31 - 100 70 18 - 59 42 12 - 40 29 Answer: a) det( A - λI ) = - λ 3 + 2 λ 2 + λ - 2 = - ( λ - 1)( λ - 2)( λ + 1) so the eigenvalues are λ 1 = 1 and λ 2 = 2 and λ 3 = - 1. To find the eigenvector for λ 1 = 1 we must solve the homegeneous equation with matrix - 1 - 1 1 1 - 1 2 2 0 1 . The matrix reduces to - 1 - 1 1 0 - 2 3 0 0 0 and the eigenvector is x 1 = 1 - 3 2 .
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