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301_F10_hw5_soln

# 301_F10_hw5_soln - AME 301 Homework#5 Solutions | Fall 2010...

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AME 301 Homework #5 Solutions | Fall 2010 1. Section 8.1 #5 A = ± 5 ± 2 9 ± 6 ² det( A ± ± I ) = (5 ± ± )( ± 6 ± ± ) ± ( ± 2)(9) = ± 2 + ± ± 12 = ( ± + 4)( ± ± 3) = 0 ± 1 = ± 4 ; ± 2 = 3 Apply checks: ± 1 ± 2 = det( A ) = ± 12 (yes) ± 1 + ± 2 = trace( A ) = ± 1 (yes) ± 1 = ± 4 : ( A ± ± 1 I ) x = ± 9 ± 2 9 ± 2 ²± x 1 x 2 ² = 0 ; x 1 = ² ± 2 9 ² : ± 2 = 3 : ( A ± ± 2 I ) x = ± 2 ± 2 9 ± 9 ²± x 1 x 2 ² = 0 ; x 2 = ³ ± 1 1 ² : x T 1 x 2 = ³ 2 9 ´ ± 1 1 ² = 11 6 = 0 eigenvectors are not orthogonal 2. The projection matrix P is P = ± cos 2 ´ cos ´ sin ´ cos ´ sin ´ sin 2 ´ ² For ´ = 36 : 87 ± , we have cos ´ = 0 : 8 and sin ´ = 0 : 6. Then P = ± 0 : 64 0 : 48 0 : 48 0 : 36 ² det( P ± ± I ) = (0 : 64 ± ± )(0 : 36 ± ± ) ± (0 : 48)(0 : 48) = ± 2 ± ± = ( ± ± 1) ± = 0 ± 1 = 1 ; ± 2 = 0 Apply checks: ± 1 ± 2 = det( P ) = 0 (yes) ± 1 + ± 2 = trace( P ) = 1 (yes) 1

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± 1 = 1 : ( P ± ± 1 I ) x = ± ± 0 : 36 0 : 48 0 : 48 ± 0 : 64 ²± x 1 x 2 ² = 0 ; x 1 = ² ± 0 : 48 0 : 36 ² : ± 2 = 0 : ( P ± ± 2 I ) x = ± 0 : 64 0 : 48 0 : 48 0 : 36 ²± x 1 x 2 ² = 0 ; x 2 = ³ ± ± 0 : 36 0 : 48 ² : x T 1 x 2 = ³ 0 : 48 0 : 36 ´ ± ± 0 : 36 0 : 48 ² = 0 eigenvectors are orthogonal 3. Projection matrix for a general value of ´ . Use shorthand c = cos ´ and s = sin ´ . P = ± cos 2 ´ cos ´ sin ´ cos ´ sin ´ sin 2 ´ ² = ± c 2 cs cs s 2 ² det( P ± ± I ) = ( c 2 ± ± )( s 2 ± ± ) ± ( cs )( cs ) = ( ± 2 ± ± ( c 2 + s 2 ) + c 2 s 2 ) ± ( cs )( cs ) = ± 2 ± ± = ( ± ± 1) ± = 0 where the coe±cient of ± in the characteristic equation was simpli²ed using c 2 + s 2 = 1. ± 1 = 1 ; ± 2 = 0 Apply checks: ± 1 ± 2 = det( P ) = 0 (yes) ± 1 + ± 2 = trace( P ) = 1 (yes) ± 1 = 1 : ( P ± ± 1 I ) x = ± c 2 ± 1 cs cs s 2 ± 1 ²± x 1 x 2 ² = ± ± s 2 cs cs ± c 2 ²± x 1 x 2 ² = 0 ; x 1 = ²
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301_F10_hw5_soln - AME 301 Homework#5 Solutions | Fall 2010...

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