midterm_F07_solns

midterm_F07_solns - MATH 202 Midterm Exam(90 minutes 1(20...

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MATH 202 - Midterm Exam October 24, 2007 (90 minutes) 1. (20 points) A = 1 2 1 1 1 2 4 1 5 0 3 6 2 6 6 0 0 2 - 6 7 row reduces to rref ( A ) = 1 2 0 4 0 0 0 1 - 3 0 0 0 0 0 1 0 0 0 0 0 (a) Find a basis for the kernel of A . - 2 1 0 0 0 , - 4 0 3 1 0 (b) Find a basis for the image of A . 1 2 3 0 , 1 1 2 2 , 1 0 6 7 (c) The vector b = 3 3 11 9 is in the image of A . Find its coordinates with respect to your basis in part (b). We can calculate: 1 1 1 3 2 1 0 3 3 2 6 11 0 2 7 9 ... 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 (Or one might notice that the basis vectors in (b) sum to the given vector.) So the coordinate vector is 1 1 1 . (d) Find the general solution to the linear system A x = b where b is the vector in part (c). Based on the calculation in (c) we know that the given vector b is the sum of columns 1, 3 and 5 of A so one particular solution is 1 0 1 0 1 . To this we add the kernel of A to get
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the general solution: x = 1 0 1 0 1 + s - 2 1 0 0 0 + t - 4 0 3 1 0 for any s,t R . Alternatively, based on our earlier calculations we know that [ A | b ] reduces to 1 2 0 4 0 1 0 0 1 - 3 0 1 0 0 0 0 1 1 0 0 0 0 0 0 So the general solution is x 1 = 1 - 2 x 2 - 4 x 4 , x 3 = 1 + 3 x 4 , x 5 = 1 , and x 2 ,x 4 are free variables.
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