Answer
$(10,3,-2)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\rightarrow \left[\begin{array}{cccc}
2 & -3 & -1 & 13 \\
-1 & 2 & -5 & 6 \\
5 & -1 & -1 & 49
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+R_{2}.\\
.\\
R_{3}\leftarrow R_{3}+5R_{2}.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & -1 & -6 & 19 \\
-1 & 2 & -5 & 6 \\
0 & 9 & -26 & 79
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{2}+R_{1}.\\
.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & -1 & -6 & 19 \\
0 & 1 & -11 & 25 \\
0 & 9 & -26 & 79
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+R_{2}.\\
.\\
R_{3}\leftarrow R_{3}-9R_{2}.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & -17 & 44 \\
0 & 1 & -11 & 25 \\
0 & 0 & 73 & -146
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}/73.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & -17 & 44 \\
0 & 1 & -11 & 25 \\
0 & 0 & 1 & -2
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+17R_{3}.\\
R_{2}\leftarrow R_{2}+11R_{3}.\\
.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 0 & 10 \\
0 & 1 & 0 & 3 \\
0 & 0 & 1 & -2
\end{array}\right]$
The solution is $(10,3,-2)$
