Each row represents and equation and the first column

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Unformatted text preview: ù é 1 -2 1 3ù 1 ê é 1 -2 ú 5 ê0 7 -5 -7 ú R2 ê0 1-1ú ê ú7 ê 7 ú ê0 3 -5 -3 ú ® ê ë û 3 -5 -3ú ë0 û So, we got a fraction showing up here. That will happen on occasion so don’t get all that excited about it. The next step is to change the 3 below this new 1 into a 0. Note that we aren’t going to bother with the -2 above it quite yet. Sometimes it is just as easy to turn this into a 0 in the same step. In this case however, it’s probably just as easy to do it later as we’ll see. © 2007 Paul Dawkins 331 http://tutorial.math.lamar.edu/terms.aspx College Algebra So, using the third row operation we get, é ù ê 1 -2 1 -2 1 3ù 1 3ú é ú ê ú R - 3R ® R ê 5 5 2 3ê ê0 ú3 1-1 0 1-1ú ê ú ® 7 7 ê ú ê ú ê0 ú 20 3 -5 -3û ë ê0 0 0ú 7 ë û Next, we need to get the number in the bottom right corner into a 1. We can do that with the second row operation. é ù ê 1 -2 1 3ú 1 3ù é 1 -2 ê ú7 ê ú -R 5 5 ê0 1-1ú 20 3 ê 0 1-1ú ê ú 7 7 ê ú ê ú ® ê0 0 1 0ú 20 ë û ê0 0 0ú 7 ë û Now, we need zeroes above this new 1. So, usin...
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