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June 28th, 2017, 01:15 PM   #1
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Solving a System of Equations with Gauss-Jordan Method

I'm a little stuck on a problem using the Gauss-Jordan method of matrices. The problem asks to solve the following set of equations:

x + 2y + 3z = 2
2x + 2y - 3z =27
3x + 2y + 5z = 10

I put into an augmented matrix:

1 2 3 = 2
2 2 -3 = 27
3 2 5 = 10

So my goal was to get all the 0's first then make the coefficients 1.

Step 1: (Make 2 and 3 0 in R2 ad R3 in the first column)
-2R1 + R2 -> R2
-3R1 + R3 -> R3

1 2 3 = 2
0 -2 -9 = 23
0 -4 -4 = 4

This is where I'm stuck. I want to make 2 and -4 0 in column 2 R1 and R3. Can I do R1 +R2 -> R1 since -2 and 2 cancel each other out? This is not a hard matrix but I'm a little tripped. I did get a little further but I got confused. Here is the rest of my math:

Step 2: Make 2 and -4 0 in col 2 R1 and R3
R1 + R2 -> R1
-2R2 + R3 -> R3

1 0 -6 = 25
0 -2 -9 = 23
0 0 14 = 42

Step 3: Make 6 and 14 0 in col 3 R1 and R2
-9R1 + 6R -> R1
14R2 + 9R3 -> R2

I also got confused here because my zero was replaced with a number like this:

-9 -12 0 = -87

Any help would be greatly appreciated. Thank you.
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June 28th, 2017, 04:31 PM   #2
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your third matrix, starting from the augmented matrix, should read:
1 0 -6 | 25
0 -2 -9 | 23
0 0 14 | -42

Last edited by mgho; June 28th, 2017 at 04:35 PM.
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June 28th, 2017, 06:24 PM   #3
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Ok thank you!
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