1 f a b 17 20 10 13 b correct 2 f a b 17 20 10 13 b 3

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1. f ( A ) = b 17 20 10 13 B correct 2. f ( A ) = b 17 20 10 13 B 3. f ( A ) = b 17 20 10 13 B 4. f ( A ) = b 17 20 10 13 B Explanation: If A can be diagonalized by A = PDP - 1 = P b d 1 0 0 d 2 B P - 1 , then f ( A ) = Pf ( D ) P - 1 = P b f ( d 1 ) 0 0 f ( d 2 ) B P - 1 . Now A can be diagonalized if we can ±nd an eigenbasis of R 2 of eigenvectors v 1 , v 2 of A corresponding to eigenvalues λ 1 , λ 2 , for then: A = P b λ 1 0 0 λ 2 B P - 1 , P = [ v 1 v 2 ] . But det[ A λI ] = v v v v 3 λ 4 2 3 λ v v v v = 8 (3 λ )(3 + λ ) = λ 2 1 = 0 , i.e. , λ 1 = 1 and λ 2 = 1. Corresponding eigenvectors are v 1 = b 2 1 B , v 2 = b 1 1 B , so P = b 2 1 1 1 B , P - 1 = b 1 1 1 2 B . Thus f ( A ) = b 2 1 1 1 Bb f (1) 0 0 f ( 1) Bb 1 1 1 2 B . Now f (1) = 2 x 3 + 3 x 2 + 3 x 1 v v v x =1 = 7 , while f ( 1) = 2 x 3 + 3 x 2 + 3 x 1 v v v x = - 1 = 3 . Consequently, f ( A ) = b 2 1 1 1 Bb 7 0 0 3 Bb 1 1 1 2 B = b 17 20 10 13 B . 014 10.0 points
kim (hk9264) – HW10 – gilbert – (53525) 8 Using the fact that e x = 1 + x + 1 2! x 2 + . . . + 1 n ! x n + . . . , compute e tA as a matrix-valued function of t when A = b 7 8 4 5 B . 1. e tA = ± 2 e 3 t e - t 2( e - t e 3 t ) e 3 t e - t 2 e - t e 3 t ² cor- rect 2. e tA = ± 2 e - t e 3 t 2( e 3 t e - t ) e - t e 3 t 2 e 3 t e - t ² 3. e tA = ± 2 e - t + e - t 2( e 3 t e - t ) e - t e 3 t 2 e 3 t + e - t ² 4. e tA = ± 2 e 3 t + e - t 2( e - t e 3 t ) e 3 t e - t 2 e - t + e 3 t ² Explanation: If A can be diagonalized by A = PDP - 1 = P b d 1 0 0 d 2 B P - 1 , then e tA = Pe tD P - 1 = P b e td 1 0 0 e td 2 B P - 1 . Now A can be diagonalized if we can Fnd an eigenbasis of R 2 of eigenvectors v 1 , v 2 of A corresponding to eigenvalues λ 1 , λ 2 , for then: A = P b λ 1 0 0 λ 2 B P - 1 , P = [ v 1 v 2 ] . But det[ A λI ] = v v v v 7 λ 8 4 5 λ v v v v = λ 2 λ 3 = ( λ 3)( λ + 1) = 0 , i.e. , λ 1 = 3 and λ 2 = 1. Corresponding eigenvectors are v 1 = b 2 1 B , v 2 = b 1 1 B , so P = b 2 1 1 1 B , P - 1 = b 1 1 1 2 B . Consequently, e tA = b 2 1 1 1 Bb e 3 t 0 0 e - t Bb 1 1 1 2 B = ± 2 e 3 t e - t 2( e - t e 3 t ) e 3 t e - t 2 e - t e 3 t ² . 015 10.0 points ±ind a matrix P so that P b d 1 0 0 d 2 B P - 1 , d 1 d 2 is a diagonalization of the matrix A = b 2 2 1 1 B 1. P = b 1 1 2 1 B 2. P = b 1 2 1 1 B 3. P = b 2 1 1 1 B 4. P = b 2 1 1 1 B correct 5. P = b 1 2 1 1 B 6. P = b 1 1 2 1 B Explanation: To begin, we must Fnd the eigenvectors and eigenvalues of A . To do this, we will use
kim (hk9264) – HW10 – gilbert – (53525) 9 the characteristic equation, det( A λI ) = 0. That is, we will look for the zeros of the characteristic polynomial. det( A λI ) = (2 λ )( 1 λ ) + 2 = λ 2 λ = ( λ 1)( λ + 0) = 0 . So D = b λ 1 0 0 λ 2 B = b 1 0 0 0 B . Now to Fnd the eigenvectors of A , we will solve for the nontrivial solution of the charac- teristic equation by row reducing the related augmented matrices: [ A λ 1 I 0 ] = b 2 1 2 0 1 1 1 0 B = b 1 2 0 1 2 0 B b 1 2 0 0 0 0 B = u 1 = b 2 1 B , while [ A λ 2 I 0 ] = b 2 + 0 2 0 1 1 + 0 0 B = b 2 2 0 1 1 0 B b 1 1 0 0 0 0 B = u 2 = b 1 1 B .

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