# hw 9 - 642:527 SOLUTIONS ASSIGNMENT 9 FALL 2007 Section 9.9...

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642:527 SOLUTIONS: ASSIGNMENT 9 FALL 2007 Section 9.9: 2 (a) See text solution. (d) The formula is (23) of the text. We have a e 1 , e 1 A (= e 1 · e 1 ) = 14, a e 2 , e 2 A = 5, a e 3 , e 3 A = 70, a e 1 , u A = 8, a e 2 , u A = 5, and a e 3 , u A = 10, so u = (4 / 7) e 1 + e 2 + (1 / 7) e 3 . 12 (d) See text solution. (f) Summary of the Gram-Schmidt process: Given v 1 , . . ., v n we de±ne ˆ e 1 = v 1 / b v 1 b . Now, once we have de±ned ˆ e 1 , . . . ˆ e k we de±ne ˆ e k +1 by a two-step process: e k +1 = v k +1 k s i =1 a ˆ e i , v k +1 A ˆ e i , ˆ e k +1 = e k +1 b e k +1 b . So for the given vectors v 1 = (1 , 1 , 1), v 2 = (1 , 0 , 1), v 3 = (1 , 1 , 0) we have ˆ e 1 = v 1 b v 1 b = 1 r a v 1 , v 1 A v 1 = 1 3 (1 , 1 , 1); e 2 = v 2 − a ˆ e 1 , v 2 A ˆ e 1 = (1 , 0 , 1) 2 3 (1 , 1 , 1) = 1 3 (1 , 2 , 1) , ˆ e 2 = e 2 b e 2 b = 1 6 (1 , 2 , 1); e 3 = v 3 − a ˆ e 1 , v 3 A ˆ e 1 − a ˆ e 2 , v 3 A ˆ e 2 = (1 , 1 , 0) 2 3 (1 , 1 , 1) + 1 6 (1 , 2 , 1) = 1 2 (1 , 0 , 1) , ˆ e 3 = e 3 b e 3 b = 1 2 (1 , 0 , 1); Section 9.10: 2 (a) See text solution.

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## This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

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hw 9 - 642:527 SOLUTIONS ASSIGNMENT 9 FALL 2007 Section 9.9...

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