# 3.2 - Math 2940 Solutions, Fall 2011 Section 3.2 1) (a)...

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Math 2940 Solutions, Fall 2011 Section 3 . 2 1 ) (a) 1 2 2 4 6 1 2 3 6 9 0 0 1 2 3 1 2 2 4 6 0 0 1 2 3 0 0 1 2 3 1 2 2 4 6 0 0 1 2 3 0 0 0 0 0 . The pivot variables are x 1 and x 3 . The other variables are free. (b) 2 4 2 0 4 4 0 8 8 2 4 2 0 4 4 0 0 0 . The pivots are x 1 and x 2 . The free variable is x 3 . 2 ) (a) The free variables are x 2 , x 4 and x 5 . So the solutions are ( - 2 , 1 , 0 , 0 , 0), (0 , 0 - 2 , 1 , 0) and (0 , 0 , - 3 , 0 , 1). (b) The free variable is x 3 . The solution is (1 , - 1 , 1). 3 ) Any solution is a linear combinations of the special solutions given in (2) above. The null space contains only ~ 0 when there are no free variables. 4 ) (a) We have the U . Continuing, we get 1 2 0 0 0 0 0 1 2 3 0 0 0 0 0 . (b) 2 4 2 0 4 4 0 0 0 1 2 1 0 1 1 0 0 0 1 0 - 1 0 1 1 0 0 0 . The point of row operations is that they do

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## This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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3.2 - Math 2940 Solutions, Fall 2011 Section 3.2 1) (a)...

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