Well this method works by getting down to either reduced row echelon form or by getting it to reduced form with a row of zeros, so you're not actually stuck; you can use backwards substitution. Because this doesn't reduce fully, it has no determinant, so it either has no solution or is parametric (infinite solutions).

So lets look at your augmented matrix:

1 2 1 4

0 -3 -1 -4

0 0 0 0

Let your third variable become the parameter t, and then move it to the side with your augmented values:

1 2 4-t

0 -3 4+t

&nbsp

Now divide row 2 by -3 for

1 2 4-t

0 1 (-4/3)-(t/3)

Then you can multiply row 2 by -2 and add it to row 1 for

1 0 (20/3)-(t/3)

0 1 (-4/3)-(t/3)

I don't know what your variables are named, but here I'd say

x=(20/3)-(t/3)

y=(-4/3)-(t/3)

z=t

[I'm mostly answering this to see if I remember what I recently learned in linear algebra correctly if someone more masterful of this comes along and finds an error please feel free to call me out]