First if we have a row in which all the entries

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Unformatted text preview: ê ú ® ê0 4 7 ê0 0 7 41ú 21ú ë û ë û We now can divide the third row by 7 to get that the number in the lower right corner into a one. é 1 -1 -3 -16 ù 1 é 1 -1 -3 -16 ù R ê0 1 0 5ú 7 3 ê 0 1 0 5ú ê ú ê ú ê0 0 7 ú ú 21û ® ê0 0 1 3û ë ë Next, we can use the third row operation to get the -3 changed into a zero. é 1 -1 -3 -16 R + 3R3 ® R ê0 1 0 5ú 1 ê ú ® ê0 0 1 3ú ë û é 1 -1 0 -7 ù ê0 1 0 5ú ê ú ê û ë 0 0 1 3ú The final step is to then make the -1 into a 0 using the third row operation again. é 1 -1 0 -7 ù R + R2 ® R ê0 1 0 5ú 1 ê ú ® ê0 0 1 3ú ë û é 1 0 0 -2 ù ê0 1 0 5ú ê ú ê 0 0 1 3ú ë û The solution to this system is then, x = -2, y = 5, z = 3 [Return to Problems] Using Gauss-Jordan elimination to solve a system of three equations can be a lot of work, but it is often no more work than solving directly and is many cases less work. If we were to do a system of four equations (which we aren’t going to do) at that poin...
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