Answer
$x=1$
$y=2$
$z=-1$
Work Step by Step
We have to solve the system of equations:
$\begin{cases}
y+2z=0\\
-x+y=1\\
2x-y+z=-1
\end{cases}$
First identify the matrices $A,X,B$ and write the system in the form $AX=B$:
$A=\begin{bmatrix}0&1&2\\-1&1&0\\2&-1&1\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}0\\1\\-1\end{bmatrix}$
$\begin{bmatrix}0&1&2\\-1&1&0\\2&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\1\\-1\end{bmatrix}$
We have to determine the solution of the system, $X$:
$X=A^{-1}B$
Determine $A^{-1}$:
$A^{-1}=\begin{bmatrix}-1&3&2\\-1&4&2\\1&-2&-1\end{bmatrix}$
Determine the solution of the system:
$X=A^{-1}B$
$X=A^{-1}=\begin{bmatrix}-1&3&2\\-1&4&2\\1&-2&-1\end{bmatrix}\begin{bmatrix}0\\1\\-1\end{bmatrix}$
$=\begin{bmatrix}0+3-2\\0+4-2\\0-2+1\end{bmatrix}$
$=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
The solution is:
$x=1$
$y=2$
$z=-1$
