Answer
$(6,-1,-1)$
Work Step by Step
Write the augmented matrix and,
using row transformations arrive at the reduced row echelon form.
-----------
$\left[\begin{array}{llll}
1 & 2 & 0 & 4\\
0 & 1 & -1 & 0\\
1 & 3 & -2 & 5
\end{array}\right]\begin{array}{l}
.\\
.\\
-R_{1}.
\end{array}$
... clear column 1,
$\left[\begin{array}{llll}
1 & 2 & 0 & 4\\
0 & 1 & -1 & 0\\
0 & 1 & -2 & 1
\end{array}\right]\begin{array}{l}
-2R_{2}.\\
.\\
-R_{2}.
\end{array}$
... clear column 2,
$\left[\begin{array}{llll}
1 & 0 & 2 & 4\\
0 & 1 & -1 & 0\\
0 & 0 & -1 & 1
\end{array}\right]\begin{array}{l}
+2R_{3}.\\
-R_{3}.\\
\div(-1).
\end{array}$
... leading 1 in row 3,
... clear column 3,
$\left[\begin{array}{llll}
1 & 0 & 0 & 6\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & -1\\
& & &
\end{array}\right]$
Solution: $(6,-1,-1)$
