HW8-510-soln - AMS 510 HW8 Solution 3.65 100 100 102 (a)...

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AMS 510 HW8 Solution 3.65 (a) E 1 = 1 0 0 0 0 1 0 1 0 , E 2 = 1 0 0 0 3 0 0 0 1 , E 3 = 1 0 2 0 1 0 0 0 1 (b) E 1 ' = 1 0 0 0 0 1 0 1 0 , E 2 ' = 1 0 0 0 1 / 3 0 0 0 1 , E 3 ' = 1 0 2 0 1 0 0 0 1 (c) Interchange C 2 and C 3 ; Replace C 2 by 3 C 2 ; Replace C 1 by 2 C 3 C 1 (d) F 1 = 1 0 0 0 0 1 0 1 0 , F 2 = 1 0 0 0 3 0 0 0 1 , F 3 = 1 0 0 0 1 0 2 0 1 3.67 A 1 = 8 12 5 5 7 3 1 2 1 ; as det B = 0 , so B has no inverse matrix; C 1 = 29 2 17 2 7 2 5 2 3 2 1 2 3 2 1 ; D 1 = 8 3 1 5 2 1 10 4 1 3.68 A 1 = 1 1 1 ⋯ − 1 n 1 0 1 1 0 0 0 0 0 0 0 0 0 1 , B 1 = 1 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 1 3.69 (a) 1 0 0 3 1 0 2 1 1  1 1 1 0 1 1 0 0 1 (b) 1 0 0 2 1 0 3 5 1  1 3 1 0 1 3 0 0 10
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(c) 1 0 0 2 1 0 3 2 1 2 1  2 3 6 0 1 3 0 0 7 2 (d) No LU decomposition. 3.70 a X 1 = 1 1 1 (b) X 2 = 6 4 0 , X 3 = 22 16 2 , X 4 = 86 62 6 3.71 1 0 0 2 1 0 3 5 1  1 0 0 0 1 0 0 0 10  1 3 1 0 1 3
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This note was uploaded on 01/31/2011 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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HW8-510-soln - AMS 510 HW8 Solution 3.65 100 100 102 (a)...

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