Answer
$x=5$
$y=4$
$z=-1$
Work Step by Step
We have to solve the system of equations:
$\begin{cases}
x-y=1\\
6x+y+20z=14\\
y+3z=1
\end{cases}$
First identify the matrices $A,X,B$ and write the system in the form $AX=B$:
$A=\begin{bmatrix}1&-1&0\\6&1&20\\0&1&3\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}1\\14\\1\end{bmatrix}$
$\begin{bmatrix}1&-1&0\\6&1&20\\0&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\14\\1\end{bmatrix}$
We have to determine the solution of the system, $X$:
$X=A^{-1}B$
Determine $A^{-1}$:
$A^{-1}=\begin{bmatrix}-17&3&-20\\-18&3&-20\\6&-1&7\end{bmatrix}$
Determine the solution of the system:
$X=A^{-1}B$
$X=\begin{bmatrix}-17&3&-20\\-18&3&-20\\6&-1&7\end{bmatrix}\begin{bmatrix}1\\14\\1\end{bmatrix}$
$=\begin{bmatrix}-17+42-20\\-18+42-20\\6-14+7 \end{bmatrix}$
$=\begin{bmatrix}5\\4\\-1\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\4\\-1\end{bmatrix}$
The solution is:
$x=5$
$y=4$
$z=-1$
