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# 1.3 - Math 2940 Solutions Fall 2011 Section 1.3 3 The...

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Math 2940 Solutions, Fall 2011 Section 1 . 3 3 ) The equations are y 1 = B 1 , y 1 + y 2 = B 2 , y 1 + y 2 + y 3 = B 3 so substituting, we see y 2 = B 2 - B 1 and y 3 = B 3 - y 1 - y 2 = B 3 - B 1 - ( B 2 - B 1 ) = B 3 - B 2 . So y 1 y 2 y 3 = 1 0 0 - 1 1 0 0 - 1 1 B 1 B 2 B 3 . 4 ) We have to solve x 1 w 1 + x 2 w 2 + x 3 w 3 = 0. The equations are x 1 + 4 x 2 + 7 x 3 = 0 , 2 x 1 + 5 x 2 + 8 x 3 = 0 , 3 x 1 + 6 x 2 + 9 x 3 = 0 . We eliminate the varible x 1 from the last trwo equations by adding - 2 and - 3 times the first equation to these two to get - 3 x 2 - 6 x 3 = 0 , - 6 x 2 - 12 x 3 = 0 . These last two equations are multiples of one another, so the solutions to one are exactly the solutions to the other. We’ll take x 2 = 2 and x 3 = - 1. Plugging back into the first equation gives x 1 = - 1. The three vectors lie in a plane . 6 ) Clearly the first two columns of 1 3 5 1 2 4 1 1 c are independent. We need to consider x 1 1 1 1 + x 2 3 2 1 = 5 4 c .

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