1.3 - Math 2940 Solutions, Fall 2011 Section 1 . 3 3 ) The...

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Unformatted text preview: 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- 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....
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell University (Engineering School).

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1.3 - Math 2940 Solutions, Fall 2011 Section 1 . 3 3 ) The...

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