Answer
$(3,1,2)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
0 & 2 & 1 & 4\\
1 & 1 & 0 & 4\\
3 & 3 & -1 & 10
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{1}.\\
R_{3}\leftarrow R_{3}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 4\\
0 & 2 & 1 & 4\\
3 & 3 & -1 & 10
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}-3R_{1 }
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 4\\
0 & 2 & 1 & 4\\
0 & 0 & -1 & -2
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{2}+R_{3 }.\\
\times(-1).
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 4\\
0 & 2 & 0 & 2\\
0 & 0 & 1 & 2
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div 2.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 4\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 2
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-R_{2}.\\
.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 0 & 3\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 2
\end{array}\right]$
The solution is $(3,1,2)$
