We can do that with the second row operation 1 2 1 3 1

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Unformatted text preview: to minimize the work as much as possible however. So, since there is a one in the first column already it just isn’t in the correct row let’s use the first row operation and interchange the two rows. é3 -2 14 ù R1 « R2 é 1 3 1ù ê 1 3 1ú ® ê3 -2 14 ú ë û ë û The next step is to get a zero below the 1 that we just got in the upper left hand corner. This means that we need to change the red three into a zero. This will almost always require us to use third row operation. If we add -3 times row 1 onto row 2 we can convert that 3 into a 0. Here is that operation. 3 1ù é 1 3 1ù R2 - 3R1 ® R2 é 1 ê3 -2 14 ú ê0 -11 11ú ® ë û ë û Next we need to get a 1 into the lower right corner of the first two columns. This means changing the red -11 into a 1. This is usually accomplished with the second row operation. If we divide the second row by -11 we will get the 1 in that spot that we need. © 2007 Paul Dawkins 328 http://tutorial.math.lamar.edu/terms.aspx College Algeb...
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