Also the path that one person finds to be the easiest

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Unformatted text preview: stems. So, let’s take a look at a couple of systems with three equations in them. In this case the process is basically identical except that there’s going to be more to do. As with two equations we will first set up the augmented matrix and then use row operations to put it into the form, é 1 0 0 pù ê0 1 0 q ú ê ú ê0 0 1 r ú ë û Once the augmented matrix is in this form the solution is x = p , y = q and z = r . As with the two equations case there really isn’t any set path to take in getting the augmented matrix into this form. The usual path is to get the 1’s in the correct places and 0’s below them. Once this is done we then try to get zeroes above the 1’s. Let’s work a couple of examples to see how this works. © 2007 Paul Dawkins 330 http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 2 Solve each of the following systems of equations. 3x + y - 2 z = 2 (a) x - 2 y + z = 3 [Solution] 2 x - y - 3z = 3 3 x + y - 2 z = -7 (b) 2 x + 2 y + z = 9 [Solution] - x - y + 3z = 6 Solution 3x + y - 2 z = 2 (a) x - 2 y...
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