Answer
$(-2,1,3)$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
4 & -3 & 1 & -8\\
-2 & 1 & -3 & -4\\
1 & -1 & 2 & 3
\end{array}\right]\ \ \begin{array}{l}
R_{1}\leftrightarrow R_{3}.\\
.\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & -1 & 2 & 3\\
-2 & 1 & -3 & -4\\
4 & -3 & 1 & -8
\end{array}\right]\ \ \begin{array}{l}
.\\
+2R_{1}.\\
-4R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & -1 & 2 & 3\\
0 & -1 & 1 & 2\\
0 & 1 & -7 & -20
\end{array}\right]\ \ \begin{array}{l}
-R_{1}.\\
\times(-1).\\
+R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & 1 & 1\\
0 & 1 & -1 & -2\\
0 & 0 & -6 & -18
\end{array}\right]\ \ \begin{array}{l}
.\\
.\\
\div(-6).
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & 1 & 1\\
0 & 1 & -1 & -2\\
0 & 0 & 1 & 3
\end{array}\right]\ \ \begin{array}{l}
-R_{3}.\\
.+R_{3}\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & 0 & -2\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 3
\end{array}\right]$
Solution: $(-2,1,3)$
