Answer
$(-1,3,5)$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
2 & -3 & 5 & 14\\
4 & -1 & -2 & -17\\
-1 & -1 & 1 & 3
\end{array}\right]\ \ \begin{array}{l}
R_{1}\leftrightarrow R_{3}.\\
.\\
.
\end{array}$
$\left[\begin{array}{llll}
-1 & -1 & 1 & 3\\
4 & -1 & -2 & -17\\
2 & -3 & 5 & 14
\end{array}\right]\ \ \begin{array}{l}
\times(-1).\\
+4R_{1}.\\
+2R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & 1 & -1 & -3\\
0 & -5 & 2 & -5\\
0 & -5 & 7 & 20
\end{array}\right]\ \ \begin{array}{l}
.\\
.\\
-R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & 1 & -1 & -3\\
0 & -5 & 2 & -5\\
0 & 0 & 5 & 25
\end{array}\right]\ \ \begin{array}{l}
.\\
.\\
\div 5.
\end{array}$
$\left[\begin{array}{llll}
1 & 1 & -1 & -3\\
0 & -5 & 2 & -5\\
0 & 0 & 1 & 5
\end{array}\right] \ \ \begin{array}{l}
+R_{3}.\\
-2R_{3}.\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & 1 & 0 & 2\\
0 & -5 & 0 & -15\\
0 & 0 & 1 & 5
\end{array}\right] \ \ \begin{array}{l}
+\frac{1}{5}R_{2}.\\
\div(-5).\\
.
\end{array}$
$\left[\begin{array}{llll}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 5
\end{array}\right]$
Solution: $(-1,3,5)$
