It just turns out that it doesnt matter how many

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Unformatted text preview: + z = 3 2 x - y - 3z = 3 Let’s first write down the augmented matrix for this system. 1 -2 2 ù é3 ê 1 -2 1 3ú ê ú ê 2 -1 -3 3ú ë û As with the previous examples we will mark the number(s) that we want to change in a given step in red. The first step here is to get a 1 in the upper left hand corner and again, we have many ways to do this. In this case we’ll notice that if we interchange the first and second row we can get a 1 in that spot with relatively little work. 1 -2 2 ù 1 3ù é3 é 1 -2 R1 « R2 ê ê 1 -2 1 3ú 1 -2 2 ú ê ú ® ê3 ú ê 2 -1 -3 3ú ê 2 -1 -3 3ú ë û ë û The next step is to get the two numbers below this 1 to be 0’s. Note as well that this will almost always require the third row operation to do. Also, we can do both of these in one step as follows. 1 3ù R2 - 3R1 ® R2 é 1 -2 1 3ù é 1 -2 ê3 ú R - 2 R ® R ê 0 7 -5 -7 ú 1 -2 2 ú 3 1 3ê ê ú ê 2 -1 -3 3ú ê0 ® 3 -5 -3ú ë û ë û Next we want to turn the 7 into a 1. We can do this by dividing the second row by 7. 1 3...
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