My Math Forum Solving a System of Equations with Gauss-Jordan Method

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 June 28th, 2017, 12:15 PM #1 Newbie   Joined: Jun 2017 From: Wisconsin Posts: 15 Thanks: 0 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.
 June 28th, 2017, 03:31 PM #2 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 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 03:35 PM.
 June 28th, 2017, 05:24 PM #3 Newbie   Joined: Jun 2017 From: Wisconsin Posts: 15 Thanks: 0 Ok thank you!

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