Answer
$x=-18$
$y=79$
$z=-27$
Work Step by Step
First identify the matrices $A,X,B$ and write the system in the form $AX=B$:
$A=\begin{bmatrix}1&1&2\\0&1&3\\3&0&-2\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}7\\-2\\0\end{bmatrix}$
$\begin{bmatrix}1&1&2\\0&1&3\\3&0&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-2\\0\end{bmatrix}$
We have to determine the solution of the system, $X$:
$X=A^{-1}B$
Determine $A^{-1}$:
$[A|I_3]=\begin{bmatrix}1&1&2&|&1&0&0\\0&1&3&|&0&1&0\\3&0&-2&|&0&0&1\end{bmatrix}$
Add $-3R_3$ to $R_1$:
$\begin{bmatrix}1&1&2&|&1&0&0\\0&1&3&|&0&1&0\\0&-3&-8&|&-3&0&1\end{bmatrix}$
Add $-R_2$ to $R_1$ and $3R_2$ to $R_3$:
$\begin{bmatrix}1&0&-1&|&1&-1&0\\0&1&3&|&0&1&0\\0&0&1&|&-3&3&1\end{bmatrix}$
Add $R_3$ to $R_1$ and $-3R_3$ to $R_2$:
$\begin{bmatrix}1&0&0&|&-2&2&1\\0&1&0&|&9&-8&-3\\0&0&1&|&-3&3&1\end{bmatrix}$
$A^{-1}=\begin{bmatrix}-2&2&1\\9&-8&-3\\-3&3&1\end{bmatrix}$
Determine the solution of the system:
$X=A^{-1}B$
$X=A^{-1}=\begin{bmatrix}-2&2&1\\9&-8&-3\\-3&3&1\end{bmatrix}\begin{bmatrix}7\\-2\\0\end{bmatrix}$
$=\begin{bmatrix}-14-4+0\\63+16-0\\-21-6+0\end{bmatrix}$
$=\begin{bmatrix}-18\\79\\-27\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-19\\79\\-27\end{bmatrix}$
The solution is:
$x=-18$
$y=79$
$z=-27$
