301_F10_hw6_soln

301_F10_hw6_soln - AME 301 Homework #6 Solutions | Fall...

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AME 301 Homework #6 Solutions | Fall 2010 1. Section 8.1 #18 Find the eigenvalues and eigenvectors of the matrix A = 2 4 3 0 12 ± 6 3 0 9 6 3 3 5 det( A ± ± I ) = (3 ± ± ) 3 + 12[ ± 36 ± (3 ± ± )9] = ± ( ± + 9)( ± ± 9) 2 = 0 ± 1 = ± 9 ; ± 2 = 9 with algebraic multiplicity = 2 ± 1 = ± 9 : ( A ± ± 1 I ) x = 2 4 12 0 12 ± 6 12 0 9 6 12 3 5 2 4 x 1 x 2 x 3 3 5 = 0 ; which row reduces to U 1 x = 2 4 12 0 12 0 12 6 0 0 0 3 5 2 4 x 1 x 2 x 3 3 5 = 0 ; x 1 = ² 2 4 ± 1 ± 1 = 2 1 3 5 : ± 2 = 9 : ( A ± ± 2 I ) x = 2 4 ± 6 0 12 ± 6 ± 6 0 9 6 ± 6 3 5 2 4 x 1 x 2 x 3 3 5 = 0 ; which row reduces to U 2 x = 2 4 ± 6 0 12 0 ± 6 ± 12 0 0 0 3 5 2 4 x 1 x 2 x 3 3 5 = 0 ; x 2 = ³ 2 4 2 ± 2 1 3 5 : Since rank( U 2 ) = rank( A ± ± 2 I ) = 2, the eigenvector space associated with ± 2 has dimension = 1. Hence, the geometric multiplicity = 1, and we have a defective matrix (less than 3 linearly independent eigenvectors). 2. A linear elastic deformation is described by the matrix equation y = Ax , where x is the location of a point prior to the deformation and y is the location after the deformation. Consider A = ± 0 : 8 0 : 6 0 : 6 1 : 7 ² : Find the eigenvalues and eigenvectors of A . det( A ± ± I ) = (0 : 8 ± ± )(1 : 7 ± ± ) ± 0 : 36 = ( ± ± 2)( ± ± 0 : 5) = 0 1
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± 1 = 2 : 0 ; ± 2 = 0 : 5 ± 1 = 2 : 0 : ( A ± ± 1 I ) x = ± ± 1 : 2 0 : 6 0 : 6 ± 0 : 3 ²± x 1 x 2 ² = 0 ; x 1 = ² ± 1 2 ² : ± 2 = 0 : 5 : ( A ± ± 2 I ) x = ± 0 : 3 0 : 6 0 : 6 1 : 2 ²± x 1 x 2 ² = 0 ; x 2 = ³ ± ± 2 1 ² : (a) Determine the angles ´ 1 and ´ 2 for the principal directions of the deformation (in which y is parallel to x ). For ± 1 , the principal direction is ´ 1 = tan ± 1 ( x 2 =x 1 ) = tan ± 1 (2 = 1) = 63 : 43 ² .
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301_F10_hw6_soln - AME 301 Homework #6 Solutions | Fall...

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