If A = [2, 1, -3; 3, 2, 1; 1, 2, -1], find A^-1 and hence solve the following system of equations: 2 x+ y -3 z =13 3 x+2 y + z =4 x+2 y - 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 $\displaystyle \left(\begin{array}{ccc} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{array}\right)\left(\begin{array}{l}x \\ \mathrm{y} \\ \mathrm{z}\end{array}\right)=\left(\begin{array}{c}13 \\ 8\end{array}\right) \Rightarrow X=\mathrm{A}^{-1} \cdot B$ $$\mathrm{A} \quad X=B $$ $$\left(\begin{array}{l} x \\ \mathrm{y} \\ \mathrm{z} \end{array}\right)=\frac{-1}{16}\left(\begin{array}{ccc} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{array}\right)\left(\begin{array}{c} 13 \\ 4 \\ 8 \end{array}\right)=\frac{-1}{16}\left(\begin{array}{c} -16 \\ -32 \\ 48 \end{array}\right)=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right) \Rightarrow x=1, \mathrm{y}=2, \mathrm{z}=-3 $$