Alg_Complete

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Unformatted text preview: g the third row operation twice as follows will do what we need done. 5 1 3ù R2 + R3 ® R2 é 1 -2 é 1 - 2 0 3ù 7 ê ú 5 ê0 1-1ú R1 - R3 ® R1 ê0 1 0 -1ú ê ú 7 ê ú ê0 0 1 0 ú ® ë û ê0 0 1 0ú ë û Notice that in this case the final column didn’t change in this step. That was only because the final entry in that column was zero. In general, this won’t happen. The final step is then to make the -2 above the 1 in the second column into a zero. This can easily be done with the third row operation. é 1 -2 0 3ù é 1 0 0 1ù ê0 ú R1 + 2 R2 ® R1 ê0 1 0 -1ú 1 0 -1ú ê ê ú ® ê0 0 1 0 ú ê0 0 1 0ú ë û ë û So, we have the augmented matrix in the final form and the solution will be, x = 1, y = -1, z = 0 This can be verified by plugging these into all three equations and making sure that they are all satisfied. [Return to Problems] © 2007 Paul Dawkins 332 http://tutorial.math.lamar.edu/terms.aspx College Algebra 3 x + y - 2 z = -7 (b) 2 x + 2 y + z = 9 - x - y + 3z = 6 Again, the first s...
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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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