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07-3 - 18.06 Professor Johnson Quiz 1 October 3 2007...

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18.06 Professor Johnson Quiz 1 October 3, 2007 SOLUTIONS 1 (20 pts.) Find all solutions to the linear system x + 2 y + z - 2 w = 5 2 x + 4 y + z + w = 9 3 x + 6 y + 2 z - w = 14 Solution: We perform elimination on the augmented matrix: 1 2 1 - 2 5 2 4 1 1 9 3 6 2 - 1 14 1 2 1 - 2 5 0 0 - 1 5 - 1 0 0 - 1 5 - 1 1 2 1 - 2 5 0 0 - 1 5 - 1 0 0 0 0 0 , So y and w are free variables. Thus special solutions to A x = 0 are given by setting y = 1 , w = 0 and y = 0 , w = 1 respectively, i.e. s 1 = - 2 1 0 0 , s 2 = - 3 0 5 1 . Moreover, a particular solution to the system is given by setting y = w = 0, i.e. x p = 4 0 1 0 . We could have read these special and particular solutions off even more easily by performing one more elimination step to get the row-reduced echelon matrix:

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1 2 0 3 4 0 0 1 - 5 1 0 0 0 0 0 = R. Notice that the last column gives the values of the pivot variables for the particular solution, and the free columns give the values of the pivot variables in the special solutions (multiplied by - 1), as was shown in class. We conclude that the general solutions to this system are given by x = x p + c 1 s 1 + c 2 s 2 = - 2 c 1 - 3 c 2 + 4 c 1 5 c 2 + 1 c 2 , where c 1 and c 2 are arbitrary constants. 2
2 (30 pts.) In class, we learned how to do “downwards” elimination to put a matrix A in upper-triangular (or echelon) form U : not counting row swaps, we subtracted multiples of pivot rows from subsequent rows to put zeros below the pivots, corresponding to multiplying A by elimination matrices. Instead, we could do elimination “leftwards” by subtracting multiples of pivot columns from leftwards columns, again to get an upper-triangular matrix U . For example, let: A = 7 6 4 6 3 12 2 0 1 We could subtract twice the third column from the first column to eliminate the 2, so that we get zeros to the left of the “pivot” 1 at the lower right. (i) Continue this “leftwards” elimination to obtain an upper-triangular matrix U from the A above, and write U in terms of A multiplied by a sequence of matrices corresponding to each leftwards-elimination step. (ii) Suppose we followed this process for an arbitrary A (not necessarily square or invertible) to get an echelon matrix U . Which of the column space and null space, if any, are the same between A and U , and why? (iii) Is the U that we get by leftwards elimination always the same as the U we get from ordinary downwards elimination? Why or why not?

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