solutions6bd

solutions6bd - Math 33a/2, Quiz 6bd, December 6, 2007 Name:...

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Math 33a/2, Quiz 6bd, December 6, 2007 Name: UCLA ID: 1. Let A be the matrix ± 1 3 2 - 4 ² . Find a general expression for A t ± 14 7 ² , where t is an arbitrary integer. Solution. Since A is a 2 by 2 matrix, we can quickly write down its characteristic polynomial: f A ( λ ) = λ 2 - (Trace A ) λ + (det A ) = λ 2 + 3 λ - 10 = 0. This factors as ( λ + 5)( λ - 2) = 0, so the eigenvalues are - 5 , 2. To find an eigenvector for λ = - 5, we compute ker( A - ( - 5) I ) = ker ± 6 3 2 1 ² . Using the usual methods, we obtain an eigenvector ± - 1 2 1 ² . Since any nonzero multiple is also acceptable, we will instead take the eigenvector v 1 = ± - 1 2 ² . To find an eigenvector for λ = 2, we compute ker( A - 2 I ) = ker ± - 1 3 2 - 6 ² . We obtain an eigenvector v
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solutions6bd - Math 33a/2, Quiz 6bd, December 6, 2007 Name:...

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