HW7.EGM4313 - Page 329, problem 13: 13 5 2 5 4 2 5 2 8 7 4...

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Unformatted text preview: Page 329, problem 13: 13 5 2 5 4 2 5 2 8 7 4 5 2 4 Let x3 27 2 243 729 9 3 7 2 x1 8 x2 2 x3 1 3 4 x1 2 2 1 is an eigenvector. 0 0 0 5 x2 2, 2 x1 2 x2 8 x1 2, x 2 2 0 9,9,9. Page 329, problem 15: 1 0 0 0 0 1 0 0 12 0 1 4 0 12 4 1 2 1 2 2 15 1 2 5 3 0 1, 1, 5, 3 1: 0 0 0 0 0 0 0 0 12 0 0 4 0 x1 12 x 2 4 x3 0 x4 0 0 0 0 0 x1 12 x 2 4 x3 4 x4 0 0 0 0 x3 0, x 2 1 0 0 1 and eigenvecto rs 0 0 0 0 0, x 1 and x 2 arbitrary 5: 4 0 0 0 0 4 0 0 12 0 4 4 Let x 3 1 x1 3, x 4 1, x 2 3 3 is an eigenvecto r 1 1 3 3: 4 0 0 0 0 4 0 0 12 0 4 4 0 x1 12 x 2 4 x3 4 x4 0 0 Let x 3 0 0 1 x1 3, x 4 1, x 2 3 3 3 is an eigenvecto r 1 1 Page 338, problem 5: Symmetric. 6 0 0 0 0 2 2 2 2 5 4 5 2 6 6 7 6 6 6 1 1,6,6 All real. Page 338, problem 9. 001 010 100 A 100 010 001 T AA T A 00 01 10 AT I A 1 0 0 1 therefore A is orthogonal 0 01 1 0 1 0 3 2 10 1, i Page 345, problem 13: The eigenvalues of A are 4, -2, 1 by inspection with eigenvectors: 4: 0 12 21 0 6 6 0 x1 0 x2 3 x3 0 0 Letx1 0 1 x2 2, x3 3 x 1 2 3 2: 6 12 21 0 0 x1 0 0 x2 6 3 x3 0 0 Let x3 0 2 0 0 x1 3 0 x2 6 0 x3 0 0 0 0 0 1 1: 3 12 21 x x1 0, x 2 1 x 0 1 2 001 012 123 X 1 X AX 1 0 0 X 1 2 1 1 21 1 0 by Gauss-Jordan method (details skipped) 00 00 20 04 p. 345, problem 16: Eigenvalues are 0, 2, -4 with eigenvectors 1 10 1,1,0. 0 01 Let 110 110 001 X 1 X AX 00 02 00 X 0 0 4 1 1/ 2 1/ 2 0 1/ 2 0 1/ 2 0 0 1 ...
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This note was uploaded on 12/15/2011 for the course EGM 4313 taught by Professor Mei during the Fall '08 term at University of Florida.

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