Answer
$x=10$
$y=3$
$z=-2$
Work Step by Step
$ \begin{bmatrix}
2 & -3 & -1 & 13\\
-1 & 2 & -5 & 6\\
5 & -1 & -1 & 49
\end{bmatrix} $
Interchange the 1st and the 2nd row and divide the new 1st row by -1.
$ \begin{bmatrix}
1& -2 & 5 & -6\\
2 & -3 & -1 & 13\\
5 & -1 & -1 & 49
\end{bmatrix} $
Add -2 times the 1st row to the 2nd row to produce a new 2nd row.
Add -5 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1& -2 & 5 & -6\\
0 & 1 & -11 & 25\\
0 & 9 & -26 & 79
\end{bmatrix} $
Add -9 times the 2nd row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1& -2 & 5 & -6\\
0 & 1 & -11 & 25\\
0 & 0 & 73 & -146
\end{bmatrix} $
Divide the last row by 73.
$ \begin{bmatrix}
1& -2 & 5 & -6\\
0 & 1 & -11 & 25\\
0 & 0 & 1& -2
\end{bmatrix} $
Use back-substitution to find the solution.
$z=-2$
$y-11z=25\rightarrow y=3$
$x-2y+5z=-6 \rightarrow x=10$
