# Solve linear system using augmented matrix

## Homework Statement

Solve using an augmented matrix:

3r-2s+4t=6
2r+3s-5t=-8
5r-4s+3t=7

2. The attempt at a solution

[3 -2 4 | 6] => [1 0 0 | x]
[2 3 -5 |-8] => [0 1 0 | y]
[5 -4 3 | 7] => [0 0 1 | z]

I have successfully used 2x2 augmented matrices, but with 3x3 it seems like there are a bunch of different options and each one just messes up everything I did to get a zero or one in the right spot. I'm not sure if there is a default starting point with these or you are looking for something to cancel to a zero or what. Thanks!

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rock.freak667
Homework Helper
Use the main diagonal as your pivot points i.e. use that row to make the other rows zero. Ex. take the 3 in the first row to make the other two elements below it zero. Repeat with the 3 in the second row and so on.

Thanks rock.freak667, I have followed your directions and that helped a lot. However, I seem to be going wrong somewhere. I solved the system of equations using software and got r=1.376E-14 , s=-1 , t=1 . When I solved using the matrix I get some fractional numbers (z=63/23). I was extra careful and double-checked all my numbers. Am I wrong?

rock.freak667
Homework Helper
Thanks rock.freak667, I have followed your directions and that helped a lot. However, I seem to be going wrong somewhere. I solved the system of equations using software and got r=1.376E-14 , s=-1 , t=1 . When I solved using the matrix I get some fractional numbers (z=63/23). I was extra careful and double-checked all my numbers. Am I wrong?
Could you post all your working with the augmented matrix?

HallsofIvy