# 3.6 - Math 2940 Solutions Fall 2011 Section 3 6 2 For A = 1...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 3 . 6 2 ) For A = 1 2 4 2 4 8 we can eyeball it and see that a basis for the row space is (1 , 2 , 4) and a basis for the column space is 1 2 . Since the row reduced echelon form is 1 2 4 0 0 0 the second and third variables are free so a basis for the null space is - 2 1 , - 4 1 . We have A T = 1 2 2 4 4 8 which has row reduced echelon form 1 2 0 0 0 0 so a basis for the left nullspace is- 2 1 . For B = 1 2 4 2 5 8 we can eyeball it and see that a basis the rows are independent so { (1 , 2 , 4) , (2 , 5 , 8) } is a basis. Since the rank is 2 an two independent columns will form a basis. We see 1 2 , 2 5 works. Reducing, we get 1 2 4 2 5 8 → 1 2 4 0 1 0 → 1 0 4 0 1 0 so the third variable is free and a basis for the nullspace is - 4 1 . For the left nullspace, we row reduce B T to get 1 2 2 5 4 8 → 1 2 0 1 4 8 → 1 2 0 1 0 0...
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3.6 - Math 2940 Solutions Fall 2011 Section 3 6 2 For A = 1...

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