Determine the product $\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 8\\ 5& -3& -1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]$ and use it to solve the system of
equations $\mathrm{x}-\mathrm{y}+\mathrm{z}=4,\mathrm{x}-2\mathrm{y}-2\mathrm{z}=9,2\mathrm{x}+\mathrm{y}+3\mathrm{z}=1$

let $\mathrm{A}=\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& 1\end{array}\right]\text{ and }\mathrm{B}=\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]$
$\begin{array}{l}\mathrm{A}\cdot \mathrm{B}=\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& 1\end{array}\right]\left[\begin{array}{lcc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]\\ =\left[\begin{array}{ccc}-4+4+8& 4-8+4& -4-8+12\\ -7+1+6& 7-2+3& -7-2+9\\ 5-3-2& -5+6-1& 5+6-3\end{array}\right]\\ \mathrm{AB}=\left[\begin{array}{lll}8& 0& 0\\ 0& 8& 0\\ 0& 0& 8\end{array}\right]=8\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ \Rightarrow \left(\frac{1}{8}\mathrm{A}\right)\mathrm{B}={\mathrm{I}}_3\left(\because \mathrm{AB}={\mathrm{BI}}_3\right)\\ \Rightarrow \left(\frac{1}{8}\mathrm{A}\right)\mathrm{B}\cdot B^{-1}={\mathrm{I}}_3{\mathrm{B}}^{-1}\\ \Rightarrow \frac{1}{8}\mathrm{A}={\mathrm{B}}^{-1}\Rightarrow {\mathrm{B}}^{-1}=\frac{1}{8}\mathrm{A}\\ \end{array}$
$\Rightarrow {\mathrm{B}}^{-1}=\frac{1}{8}\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& 1\end{array}\right]\dots \left(\mathrm{i}\right)$
Matrix from of equations
$\left[\begin{array}{ccc}1& -1& 1\\ 1& -2& -2\\ 2& 1& 3\end{array}\right]\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}4\\ 9\\ 1\end{array}\right]$
$\begin{array}{l}\mathrm{BX}=\mathrm{C}\\ \Rightarrow \mathrm{X}={\mathrm{B}}^{-1}\mathrm{C}\end{array}$
$\begin{array}{l}\Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}-4& 4& 4\\ -7& 1& 3\\ 5& -3& 1\end{array}\right]\left[\begin{array}{c}4\\ 9\\ 1\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{8}\left[\begin{array}{c}-16+36+4\\ -25+9+3\\ 20-27-1\end{array}\right]\\ =\frac{1}{8}\left[\begin{array}{c}24\\ -16\\ -8\end{array}\right]=\left[\begin{array}{c}3\\ -2\\ -1\end{array}\right]\end{array}$
$\text{ Hence, }\mathrm{x}=3,\mathrm{y}=-2,\mathrm{z}=-1$