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 GaussJordan 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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 Spring '12
 MrVinh
 ........., Paul Dawkins

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