Answer
$(3,1,1)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\rightarrow \left[\begin{array}{cccc}
2 & 1 & 0 & 7 \\
2 & -1 & 1 & 6 \\
3 & -2 & 4 & 11
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{3}-R_{1}.\\
R_{2}\leftarrow R_{2}-R_{1}.\\
.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & -3 & 4 & 4 \\
0 & -2 & 1 & -1 \\
3 & -2 & 4 & 11
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow-0.5R_{2}.\\
R_{3}\leftarrow R_{3}-3R_{1}.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & -3 & 4 & 4 \\
0 & 1 & -0.5 & 0.5 \\
0 & 7 & -8 & -1
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+3R_{2}.\\
.\\
R_{3}\leftarrow R_{3}-7R_{2}.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 2.5 & 5.5 \\
0 & 1 & -0.5 & 0.5 \\
0 & 0 & -4.5 & -4.5
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}/(-0.45).
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 2.5 & 5.5 \\
0 & 1 & -0.5 & 0.5 \\
0 & 0 & 1 & 1
\end{array}\right] \rightarrow\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-2.5R_{2}.\\
R_{2}\leftarrow R_{2}+0.5R_{2}\\
.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 0 & 3 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1
\end{array}\right]$
The solution is $(3,1,1)$
