Answer
$x=3$
$y=1$
$z=1$
Work Step by Step
$ \begin{bmatrix}
2 & 1 & 0 & 7\\
2 & -1 & 1 & 6\\
3 & -2 &4 & 11
\end{bmatrix} $
Divide the 1st row by 2.
$ \begin{bmatrix}
1 & \frac{1}{2} & 0 & \frac{7}{2}\\
2 & -1 & 1 & 6\\
3 & -2 &4 & 11
\end{bmatrix} $
Add -2 times the 1st row to the 2nd row to produce a new 2nd row.
Add -3 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & \frac{1}{2} & 0 & \frac{7}{2}\\
0 & -2 & 1 & -1\\
0 & -\frac{7}{2} &4 & \frac{1}{2}
\end{bmatrix} $
Divide the second row by -2.
$ \begin{bmatrix}
1 & \frac{1}{2} & 0 & \frac{7}{2}\\
0 & 1 & -\frac{1}{2} & \frac{1}{2}\\
0 & -\frac{7}{2} &4 & \frac{1}{2}
\end{bmatrix} $
Add $\frac{7}{2}$ times the 2nd row to the 3rd row.
$ \begin{bmatrix}
1 & \frac{1}{2} & 0 & \frac{7}{2}\\
0 & 1 & -\frac{1}{2} & \frac{1}{2}\\
0 & 0 &\frac{9}{4} & \frac{9}{4}
\end{bmatrix} $
Use back-substitution to find the solution.
$\frac{9}{4}z=\frac{9}{4} \rightarrow z=1$
$-2y+z=-1\rightarrow y=1$
$2x+y=7 \rightarrow x=3$
