Answer
$x=1$
$y=2$
$z=-1$
Work Step by Step
We have to solve the system of equations:
$\begin{cases}
3x-2y+z=-2\\
4x-5y+3z=-9\\
2x-y+5z=-5
\end{cases}$
First identify the matrices $A,X,B$ and write the system in the form $AX=B$:
$A=\begin{bmatrix}3&-2&1\\4&-5&3\\2&-1&5\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}-2\\-9\\-5\end{bmatrix}$
$\begin{bmatrix}3&-2&1\\4&-5&3\\2&-1&5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2\\-9\\-5\end{bmatrix}$
We have to determine the solution of the system, $X$:
$X=A^{-1}B$
Determine $A^{-1}$:
$A^{-1}=\begin{bmatrix}\dfrac{11}{16}&-\dfrac{9}{32}&\dfrac{1}{32}\\\dfrac{7}{16}&-\dfrac{13}{32}&\dfrac{5}{32}\\-\dfrac{3}{16}&\dfrac{1}{32}&\dfrac{7}{32}\end{bmatrix}$
Determine the solution of the system:
$X=A^{-1}B$
$X=\begin{bmatrix}\dfrac{11}{16}&-\dfrac{9}{32}&\dfrac{1}{32}\\\dfrac{7}{16}&-\dfrac{13}{32}&\dfrac{5}{32}\\-\dfrac{3}{16}&\dfrac{1}{32}&\dfrac{7}{32}\end{bmatrix}\begin{bmatrix}-2\\-9\\-5\end{bmatrix}$
$=\begin{bmatrix}-\dfrac{22}{16}+\dfrac{81}{32}-\dfrac{5}{32}\\-\dfrac{14}{16}+\dfrac{117}{32}-\dfrac{25}{32}\\\dfrac{6}{16}-\dfrac{9}{32}-\dfrac{35}{32}\end{bmatrix}$
$=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
The solution is:
$x=1$
$y=2$
$z=-1$
