Mathlamaredutermsaspx college algebra two of the

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Unformatted text preview: o do. Example 2 Solve the following system of equations using augmented matrices. 3 x - 3 y - 6 z = -3 2 x - 2 y - 4 z = 10 -2 x + 3 y + z = 7 Solution Here’s the augmented matrix for this system. é 3 - 3 - 6 - 3ù ê 2 -2 -4 10 ú ê ú ê -2 3 1 7ú ë û We can get a 1 in the upper left corner by dividing by the first row by a 3. é 3 -3 -6 -3ù 1 é 1 -1 -2 -1ù ê 2 -2 -4 10 ú R1 ê 2 -2 -4 10 ú ê ú3 ê ú ê ú ê ú 3 1 7 û ® ë -2 3 1 7û ë -2 Next we’ll get the two numbers under this one to be zeroes. é 1 -1 -2 -1ù R2 - 2 R1 ® R2 é 1 -1 -2 -1ù ê 2 -2 -4 10ú R + 2 R ® R ê0 0 0 12 ú 1 3ê ê ú3 ú ê -2 ê0 1 -3 5ú 3 1 7ú ® ë û ë û And we can stop. The middle row is all zeroes except for the final entry which isn’t zero. Note that it doesn’t matter what the number is as long as it isn’t zero. Once we reach this type of row we know that the system won’t have any solutions and so there isn’t any reason to go any farther. Okay, let’s see how we solve a system of three equations with an infinity number...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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