Answer
$(1,0,1)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
1 & 1 & 1 & 2\\
2 & -3 & 2 & 4\\
4 & 1 & -3 & 1
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{2}-2R_{1}.\\
R_{3}\leftarrow R_{3}-4R_{1}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 1 & 2\\
0 & -5 & 0 & 0\\
0 & -3 & -7 & -7
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow-\frac{1}{5}R_{2}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 1 & 2\\
0 & 1 & 0 & 0\\
0 & -3 & -7 & -7
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}- R_{2}.\\
.\\
R_{3}\leftarrow R_{3}+3R_{2}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 1 & 2\\
0 & 1 & 0 & 0\\
0 & 0 & -7 & -7
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow-\frac{1}{7}R_{3}
\end{array}\right) $
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 1 & 2\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-R_{3}.\\
.\\
\end{array}\right) $
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 1
\end{array}\right]$
The solution is $(1,0,1)$
