Note that if wed gotten 1 1 6 0 1 0 we would have been

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Unformatted text preview: tep is to write down the augmented matrix. é 3 1 -2 -7 ù ê2 2 1 9ú ê ú ê -1 -1 3 6 ú ë û We can’t get a 1 in the upper left corner simply by interchanging rows this time. We could interchange the first and last row, but that would also require another operation to turn the -1 into a 1. While this isn’t difficult it’s two operations. Note that we could use the third row operation to get a 1 in that spot as follows. é 3 1 -2 -7 ù é 1 -1 -3 -16 ù ê2 2 ú R1 - R2 ® R1 ê 2 2 1 9ú 1 9ú ê ê ú ® ê -1 -1 3 6 ú ê -1 -1 3 6ú û ë ë û Now, we can use the third row operation to turn the two red numbers into zeroes. é 1 -1 -3 -16 ù R2 - 2 R1 ® R2 é 1 -1 -3 -16 ù ê2 2 1 9 ú R3 + R1 ® R3 ê0 4 7 41ú ê ú ê ú ê -1 -1 3 ê0 -2 0 -10 ú 6ú ® ë û ë û The next step is to get a 1 in the spot occupied by the red 4. We could do that by dividing the whole row by 4, but that would put in a couple of somewhat unpleasant fractions. So, instead of doing that we are going to interchange the second and third row. The reason for this will be...
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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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