Find the inverse of the matrix $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$. Using the inverse, $\displaystyle \mathrm{A}^{-1}$, solve the system of linear equations \[x-y+2 z=1 ; 2 y-3 z=1 ; 3 x-2 y+4 z=3 . \]

$$\mathrm{A}=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] $\displaystyle \therefore \mathrm{A}$ is invertible. \begin{aligned} & \operatorname{adj} \mathrm{A}=\left[\begin{array}{rrr} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right] \\ & \Rightarrow \mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|} \operatorname{adj} \mathrm{A}=\left[\begin{array}{rrr} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right] \end{aligned} The given system of equation can be written as $\displaystyle \mathrm{AX}=\mathrm{B}$, when \mathrm{A}=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right], X=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1 \\ 1 \\ \end{array}\right] $$