Answer
$(-2,3,3)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
1 & 1 & 1 & 4\\
-1 & 2 & 3 & 17\\
2 & -1 & 0 & -7
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{2}+R_{1}.\\
R_{3}\leftarrow R_{3}-2R_{1}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & -3 & -2 & -15
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}+R_{2}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & 0 & 2 & 6
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow\frac{1}{2}R_{3}
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 1 & 4\\
0 & 3 & 4 & 21\\
0 & 0 & 1 & 3
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-R_{3}.\\
R_{2}\leftarrow R_{2}-4R_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 1\\
0 & 3 & 0 & 9\\
0 & 0 & 1 & 3
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{3}\leftarrow\frac{1}{3}R_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 1 & 0 & 1\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 3
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-R_{2}.\\
.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 0 & -2\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 3
\end{array}\right]$
The solution is $(-2,3,3)$
