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## 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!