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Unformatted text preview: AMS510 Analytical Method for AMS December 11, 2011 Midterm Exam III You are encouraged to do all the problems, the best 6 will be counted [(1).] Find the solution to the equation Ax = b , if exists, where A and b are given by: A = 2 1 2 1 4 1 5 0 4 2 1 0 1 1 2 2 b = 3 5 1 Solution: We can simply the augmented matrix to get the solution 2 1 2 1 4 1 5 3 4 2 1 0 5 1 1 2 2 r 1 2 r 4 3 6 3 4 1 5 3 4 2 1 5 1 1 2 2 r 3 r 2 1 2 1 4 1 5 3 1 6 8 1 1 2 2 r 2 4 r 4 1 2 1 5 13 8 3 1 6 8 1 1 2 2 r 2 5 r 1 r 3 r 1 1 2 1 3 3 3 8 1 8 1 1 2 2 r 2 1 / 3 r 4 + r 1 0 1 2 1 0 0 1 1 1 0 0 8 1 8 1 0 1 r 3 +8 r 2 0 1 2 1 0 0 0 1 1 1 0 0 0 7 0 1 0 0 1 r 3 * ( 1 / 7) exchange rows 1 0 0 1 0 1 2 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 0 0 2 0 0 1 0 1 0 0 0 1 Therefore the solution is (0 , 2 , 1 , 0) T . 2 [(2).] Consider linear mapping F : v u = Av , where A = 2 1 2 1 4 1 5 0 4 2 4 1 Find the dimension and basis for the kernel and the image of F . Solution: Reduce the matrix A to echelon form 2 1 2 1 4 1 5 0 4 2 4 1 2 1 2 1 1 1 2 1 We can see there is only one free variable z . Set it to be 1 and we will get a basis for ker ( F ) { ( 3 / 2 , 1 , 1 , 0) T } . And dim ( ker ( F )) = 1 ....
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This note was uploaded on 02/28/2012 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Feinberg,E

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