PS #9 SOLUTION - 1 dna 31 = .3 = 2 1 seulavnegie owt evah...

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Math 3013 Homework Set 8 Problems from § 5.1 (pgs. 300-301 of text): 2,4,6,8,10,12,17,19,23 1. (Problems 5.1.2, 5.1.4, 5.1.6, 5.1.8, 5.1.10, and 5.1.12 in text). Find the characteristic polynomial, the real eigenvalues, and the corresponding eigenvectors for the following matrices. (a) A = 75 - 10 - 8 The characteristic polynomial is P ( λ ) = det ( A - λ I )= 7 - λ 5 - 10 - 8 - λ =(7 - λ )( - 8 - λ ) - (5) ( - 10) = λ 2 + λ - 6=( λ +3)( λ - 2) The eigenvalues of A correspond to the roots of P ( λ ) = 0; so we have two eigenvalues λ 1 = - 3 and λ 2 =2. The eigenspace corresponding to the eigenvalue λ 1 = - 3 is the solution set of ( A - ( - 3) I ) x =0 ; i.e., the null space of the matrix A - ( - 3) I = 7+3 5 - 10 - 8+3 = 10 5 - 10 - 5 R 1 1 5 R 1 R 2 R 2 - R 1 ------------→ 21 00 The null space of the last matrix is the solution set of 2 x 1 + x 2 0=0 x 1 = - 1 2 x 2 x = - 1 2 x 2 x 2 span - 1 2 1 Thus, the eigenvectors corresponding to the eigenvalue λ 1 = - 3 are thus of the form v 1 = r - 1 2 1 ,r R -{ 0 } The eigenspace corresponding to the eigenvalue λ 2 = 2 is the solution set of ( A - (2) I ) x = 0; i.e., the null space of the matrix A - ( - 3) I = 7 - 25 - 10 - 8 - 2 = 55 - 10 - 10 R 1 1 5 R 1 R 2 R 2 +2 R 1 -------------→ 11 The null space of the last matrix is the solution set of x 1 + x 2 x 1 = - x 2 x = - x 2 x 2 span - 1 1 Thus, the eigenvectors corresponding to the eigenvalue λ 2 = 2 are thus of the form v 2 = r - 1 1 R 0 } (b) A = - 7 - 5 16 17 The characteristic polynomial is P ( λ )=det( A - λ I - 7 - λ - 5 16 17 - λ = λ 2 - 10 λ - 39 = ( λ - 13) ( λ +3) The eigenvalues of A correspond to the roots of P ( λ ) = 0; so we have two eigenvalues λ 1 = 13 and λ 2 = - 3. 1
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2 The eigenspace corresponding to the eigenvalue λ 1 = 13 is the solution set of ( A - (13) I ) x =0 ; i.e., the null space of the matrix A - (13) I = - 20 - 5 16 4 R 1 →- 1 5 R 1 R 2 R 2 + 4 5 R 1 -------------→ 41 00 The null space of the last matrix is the solution set of 4 x 1 + x 2 0=0 x 1 = - 1 4 x 2 x = - 1 4 x 2 x 2 span - 1 4 1 Thus, the eigenvectors corresponding to the eigenvalue λ 1 = 13 are thus of the form v 1 = r - 1 4 1 ,r R -{ 0 } The eigenspace corresponding to the eigenvalue λ 2 = - 3 is the solution set of ( A - ( - 3) I ) x ; i.e., the null space of the matrix A - ( - 3) I = - 4 - 5 16 20 R 1 R 1 R 2 R 2 +4 R 1 45 The null space of the last matrix is the solution set of 4 x 1 +5 x 2 x 1 = - 5 4 x 2 x = - 5 4 x 2 x 2 span - 5 4 1
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PS #9 SOLUTION - 1 dna 31 = .3 = 2 1 seulavnegie owt evah...

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