To solve these systems we will use either the

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Unformatted text preview: f examples. We will start out with the two systems of equations that we looked at in the first section that gave the special cases of the solutions. Example 1 Use augmented matrices to solve each of the following systems. x- y =6 (a) [Solution] -2 x + 2 y = 1 (b) 2 x + 5 y = -1 [Solution] -10 x - 25 y = 5 Solution (a) x- y =6 -2 x + 2 y = 1 Now, we’ve already worked this one out so we know that there is no solution to this system. Knowing that let’s see what the augmented matrix method gives us when we try to use it. We’ll start with the augmented matrix. é 1 -1 6ù ê -2 2 1ú ë û Notice that we’ve already got a 1 in the upper left corner so we don’t need to do anything with that. So, we next need to make the -2 into a 0. é 1 -1 6 ù R2 + 2 R1 ® R2 é 1 -1 6 ù ê -2 2 1ú ê0 0 13ú ® ë û ë û Now, the next step should be to get a 1 in the lower right corner, but there is no way to do that without changing the zero in the lower left corner. That’s a problem, because we must have a zero in that spot as well...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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