Help with Gauss-Jordan Elimination

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Hello everyone, this is my first time posting here, I found the website looking for help. For my finite math homework one of the questions ask to solve a system through the Gauss-Jordan Elimination; here's what I have so far.

Homework Statement



3x + y -2z = 2
x - 2y +z = 3
2x - y -3z = 3

Homework Equations



So when I put it in the matrix I get this:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ]
[ 2 -1 -3 3 ]

The Attempt at a Solution



My problem is that I have not really mastered the Gauss-Jordan elimination. So I don't know the exact rules as to how I can proceed with this. I don't know if there are any limitations as to how much I can multiply, substract, add rows to get to:

[ 1 0 0 ? ]
[ 0 1 0 ? ]
[ 0 0 1 ? ]

Thank you,
 
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You can multiply/divide a row by any number and add/subtract any multiple of a row to a different row.
 
Here is what I got so far:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ] -1
[ 2 -1 -3 3 ] -2

[ 3 1 -2 2 ]
[ 0 -3 0 2 ]
[ 0 -3 -5 1 ] + row 2

[ 3 1 -2 2 ] + 2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 3 0 4 ] + r2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 3 0 4 ] + r2
[ 0 -3 0 2 ]
[ 0 0 -5 3 ]

[ 5 0 0 6 ] /5
[ 0 -3 0 2 ] /-3
[ 0 0 -5 3 ] /-5

[ 1 0 0 1.2 ]
[ 0 1 0 -.666... ]
[ 0 0 1 -.6 ]

I don't know what I am doing wrong.
 
Last edited:
clickclick said:
Here is what I got so far:

[ 3 1 -2 2 ]
[ 1 -2 1 3 ] -1
[ 2 -1 -3 3 ] -2

[ 3 1 -2 2 ]
[ 0 -3 0 2 ]
[ 0 -3 -5 1 ] + row 2
You're not allowed to do that. You can't simply subtract all the entries of a row by some constant. You can only change the numbers in each row by either:
1. Adding some multiple of another row to it or
2. Multiplying that row by a constant.
 
Using Gauss-Jordan to solve linear equations is no different to if you were going to do it without matricies.

If you were solving this problem without matricies would you simply subtract or add random constants to coefficients?? No, necause that would not make sense. What you would do is to add multiples of one equation to another equation, or multiply equations by some constants (as Defennder said).

Hopefully you see why you can't simply add constants to coefficients now.

-Spoon
 

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