Find A^-1, if A = [1, 2, 1; 2, 3, -1; 1, 0, 1]. Hence, solve the following system of equations: x+2 y+z=5; 2 x+3 y=1; x-y+z=8

For Matrix $\displaystyle \mathrm{A}=\left(\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & -1 \\ 1 & 0 & 1 \end{array}\right)$, Adjoint of Matrix A is $\displaystyle |\mathrm{A}|=-6 \neq 0$ so, $\displaystyle \mathrm{A}^{-1}$ exists. $\displaystyle \operatorname{adj} \mathrm{A}=\left(\begin{array}{ccc} 3 & -2 & -5 \\ -3 & 0 & 3 \\ -3 & 2 & -1 \end{array}\right)$, Thus, $\displaystyle \mathrm{A}^{-1}=\frac{-1}{6}\left(\begin{array}{ccc} 3 & -2 & -5 \\ -3 & 0 & 3 \\ -3 & 2 & -1 \end{array}\right)$ so, Given equation can be written into a matrix equation as $\displaystyle \left(\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 1 & -1 & 1 \end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}5 \\ 8\end{array}\right) \Rightarrow \mathrm{X}=\left(\mathrm{A}^{T}\right)^{-1} \cdot \mathrm{B}=\mathrm{X}=\left(\mathrm{A}^{-1}\right)^{T} \cdot \mathrm{B}$ $$\mathrm{A}^{T} \quad \mathrm{X}=\mathrm{B} $$ $\displaystyle \left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\frac{-1}{6}\left(\begin{array}{ccc} 3 & -3 & -3 \\ -2 & 0 & 2 \\ -5 & 3 & -1 \end{array}\right)\left(\begin{array}{l}5 \\ 8\end{array}\right)=\frac{-1}{6}\left(\begin{array}{c}-12 \\ -30\end{array}\right)=\left(\begin{array}{c}2 \\ -1 \\ 5\end{array}\right)$ $\displaystyle \therefore x=2, y=-1, z=5$