Find the inverse of the matrix

\[ A= \begin{pmatrix} 3 & 0 & -2 \\ 1 & 4 & -1 \\ 2 & -3 & 0 \\ \end{pmatrix} \]

\begin{align} & We\ know\ that \\ A^{-1} & = \frac{1}{|A|}\ adj\ A \quad \quad \enclose{circle}{\color{black}{1}} \\ \\ \\ & So\ first\ find\ the\ determinant\ |A| \\ \\ |A| & = 3[4(0)-(-3)(-1)]-0[1(0)-2(-1)]-2[1(-3)-2(4)] \\ & = -9-0+22 \\ & = 13 \\ \\ & Then\ find\ A^T \quad (i.e.\ transpose\ A) \\ \\ A^T & = \begin{pmatrix} 3 & 1 & 2 \\ 0 & 4 & -3 \\ -2 & -1 & 0 \\ \end{pmatrix} \\ \\ & And\ finally,\ replace\ the\ elements\ of\ the\ transpose\ matrix\ \\ & with\ their\ cofactors\ to\ produce\ adj\ A \quad(i.e.\ adjugate\ A) \\ \\ adj\ A & =\begin{pmatrix} \begin{vmatrix} 4 & -3 \\ -1 & 0 \end{vmatrix} & -\begin{vmatrix} 0 & -3 \\ -2 & 0 \end{vmatrix} & \begin{vmatrix} 0 & 4 \\ -2 & -1 \end{vmatrix} \\ -\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix} & \begin{vmatrix} 3 & 2 \\ -2 & 0 \end{vmatrix} & - \begin{vmatrix} 3 & 1 \\ -2 & -1 \end{vmatrix} \\ \begin{vmatrix} 1 & 2 \\ 4 & -3 \end{vmatrix} & -\begin{vmatrix} 3 & 2 \\ 0 & -3 \end{vmatrix} & \begin{vmatrix} 3 & 1 \\ 0 & 4 \end{vmatrix} \\ \end{pmatrix} \\ \\ adj\ A & =\begin{pmatrix} -3 & 6 & 8 \\ -2 & 4 & 1 \\ -11 & 9 & 12 \\ \end{pmatrix} \\ \\ \\ & Substituting\ our\ results\ into \quad \enclose{circle}{\color{black}{1}} \quad gives\ A^{-1} \ (i.e.\ inverse\ A)\\ \\ A^{-1} & = \frac{1}{13} \begin{pmatrix} -3 & 6 & 8 \\ -2 & 4 & 1 \\ -11 & 9 & 12 \\ \end{pmatrix} \\ \\ \\ & To\ check\ our\ answer \\ \\ A^{-1}A & = \frac{1}{13} \begin{pmatrix} -3 & 6 & 8 \\ -2 & 4 & 1 \\ -11 & 9 & 12 \\ \end{pmatrix} \begin{pmatrix} 3 & 0 & -2 \\ 1 & 4 & -1 \\ 2 & -3 & 0 \\ \end{pmatrix} \\ \\ & = \frac{1}{13} \begin{pmatrix} 13 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \\ \end{pmatrix} \\ \\ & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}= I \quad as\ required \end{align}