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Unformatted text preview: 1.3. GaussJordan Elimination Remember though that we originally wanted to put our matrix into the ideal form 1 1 1 A B C Gaussian Elimination takes us to the matrix 1 1 2 1 2 1 3 1 4 But how do we go further? The answer is one final step: Change all the entries above the 1s into 0s. You can go from left to right, or from right to left. Going from right to left saves a little bit of computation, since you will have more 0s in the matrix. 1 1 2 1 2 1 3 1 4 1 1 2 1 2 1 3 1 4  1 2 3 R 1 = r 1 2 R 3 1 1 1 2 1 5 1 4 1 1 2 1 2 1 3 1 4  1 2 3 R 1 = r 1 2 R 3 1 1 1 2 1 5 1 4  2 2 3 R 2 = r 2 2 R 3 1 {1 1 1 5 7 4 1 1 2 1 2 1 3 1 4  1 2 3 R 1 = r 1 2 R 3 1 1 1 2 1 5 1 4  2 2 3 R 2 = r 2 2 R 3 1 1 1 1 5 7 4  1 + 2 R 1 = r 1 + R 2 1 1 1 12 7 4 Remember that getting the matrix in the form 1 1 1 12 7 4 means that weve solved the system: x = 12 , y = 7 , and z = 4 (just like before!). If we do this extra step, then we have done Gauss Jordan Elimination , and the matrix is now in reduced row echelon form (or, as the book calls it, reduced ech elon form ). Note that a matrix has only one reduced row echelon form, but can have infinitely many row echelon forms! 1 1 2 1 2 1 3 1 4 1 1 1 12 7 4 Question: Why would we want to do GaussJordan Elimination? Question: Why would we want to do GaussJordan Elimination? Answer: Because some weird things can happen in systems of linear equations! 1. x + 2 y = 2 x + 2 y = 3 2. x + 2 y = 2 2 x + 4 y = 4 Question: Why would we want to do GaussJordan Elimination? Answer: Because some weird things can happen in systems of linear equations! 1. x + 2 y = 2 x + 2 y = 3 No solutions! 2. x + 2 y = 2 2 x + 4 y = 4 Infinitely many solutions! A more interesting example is the following: 3 x + 3 y 3 z = 4 x + y + z = 1 5 x + 5 y + 8 z = 8 Lets do Gaussian Elimination on the matrix 3 3 3 1 1 1 5 5 8 4 1 8 3 3 3 1 1 1 5 5 8 4 1 8 3 3 3 1 1 1 5 5 8 4 1 8  1 2 SWAP ( R 1 ,R 2 ) 1 1 1 3 3 3 5 5 8 1 4 8 3 3 3 1 1 1 5 5 8 4 1 8  1 2 SWAP ( R 1 ,R 2 ) 1 1 1 3 3 3 5 5 8 1 4 8  2 3 1 R 2 = r 2 3 R 1 1 1 1 6 5 5 8 1 7 8 3 3 3 1 1 1 5 5 8 4 1 8  1 2 SWAP ( R 1 ,R 2 ) 1 1 1 3 3 3 5 5 8 1 4 8  2 3 1 R 2 = r 2 3 R 1 1 1 1 6 5 5 8 1 7 8  3 5 1 R 3 = r 3 5 R 1 1 1 1 6 3 1 7 3 The next step is to ignore the first row and first column, and work with the smaller matrix: 1 1 1 6 3 1 7 3 The next step is to ignore the first row and first...
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 Spring '10
 heckman
 GaussJordan Elimination, Gaussian Elimination

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