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**Unformatted text preview: **Math 2940 Solutions, Fall 2011 Section 3 . 4 3 ) 1 3 3 | 1 2 6 9 | 5- 1- 3 3 | 5 1 3 3 | 1 0 3 | 3- 1- 3 3 | 5 1 3 3 | 1 0 0 3 | 3 0 0 6 | 6 1 3 3 | 1 0 0 1 | 1 0 0 6 | 6 1 3 3 | 1 0 0 1 | 1 0 0 0 | 1 3 0 | - 2 0 0 1 | 1 0 0 0 | . The free variable is y . Setting it equal to 0 we get ~x p = (- 2 , , 1) and the typical element of the null space is of the form (- 3 s, s, 0). 5 ) 1 2- 2 | b 1 2 5- 4 | b 2 4 9- 8 | b 3 1 2- 2 | b 1 0 1 | b 2- 2 b 1 4 9- 8 | b 3 1 2- 2 | b 1 0 1 | b 2- 2 b 1 0 1 | b 3- 4 b 1 1 2- 2 | b 1 0 1 | b 2- 2 b 1 0 0 | b 3- b 2- 2 b 1 so we must have b 3- b 2- 2 b 1 = 0. Assuming this, we have 1 2- 2 | b 1 0 1 | b 2- 2 b 1 0 0 | 1 0- 2 | - 2 b 2 + 5 b 1 0 1 | b 2- 2 b 1 0 0 | The free variable is z so ~x p = (5 b 1- 2 b 2 , b 2- 2 b 1 , 0) and the typical element of the null space...

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