If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1\end{array}\right]$, find $\displaystyle \mathrm{A}^{-1}$ and hence solve the following system of equations : $\displaystyle 2 x+\mathrm{y}-3 \mathrm{z}=13$ $\displaystyle 3 x+2 \mathrm{y}+\mathrm{z}=4$ $\displaystyle x+2 \mathrm{y}-\mathrm{z}=8$

For Matrix $\displaystyle \mathrm{A}=\left(\begin{array}{ccc} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{array}\right),|\mathrm{A}|=-16 \neq 0$ so, $\displaystyle \mathrm{A}^{-1}$ exists.\operatorname{adj} \mathrm{A}=\left(\begin{array}{ccc} -4 & -5 & 7 4 & 1 & -11 4 & -3 & 1 \end{array}\right),Thus, $\displaystyle \mathrm{A}^{-1}=\frac{-1}{16}\left(\begin{array}{ccc} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{array}\right)$ so, Given equation can be written into a matrix equation as $\displaystyle \left(\begin{array}{ccc} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{array}\right)\left(\begin{array}{l}x \\ \mathrm{y} \\ \mathrm{z}\end{array}\right)=\left(\begin{array}{c}13 \\ 8\end{array}\right) \Rightarrow X=\mathrm{A}^{-1} \cdot B$\mathrm{A} \quad X=B \left(\begin{array}{l} x \mathrm{y} \mathrm{z} \end{array}\right)=\frac{-1}{16}\left(\begin{array}{ccc} -4 & -5 & 7 4 & 1 & -11 4 & -3 & 1 \end{array}\right)\left(\begin{array}{c} 13 4 8 \end{array}\right)=\frac{-1}{16}\left(\begin{array}{c} -16 -32 48 \end{array}\right)=\left(\begin{array}{c} 1 2 -3 \end{array}\right) \Rightarrow x=1, \mathrm{y}=2, \mathrm{z}=-3 $$