Given,
The matrix $\mathrm{A}=\left[\begin{array}{ccc}2& -3& 3\\ 2& 2& 3\\ 3& -2& 2\end{array}\right]$

The determinant of A, |A| = 2(4 + 6) + 3(4 – 9) + 3( – 4 – 6) = 20 – 15 – 30 = – 25 ≠ 0

This indicates that the matrix have inverse

In order to find adjoint of the matrix A let's find the minors of the matrix A $=\left[\begin{array}{ccc}\left|\begin{array}{cc}2& 3\\ -2& 2\end{array}\right|& \left|\begin{array}{ll}2& 3\\ 3& 2\end{array}\right|& \left|\begin{array}{cc}2& 2\\ 3& -2\end{array}\right|\\ \left|\begin{array}{cc}-3& 3\\ -2& 2\end{array}\right|& \left|\begin{array}{ll}2& 3\\ 3& 2\end{array}\right|& \left|\begin{array}{cc}2& -3\\ 3& -2\end{array}\right|\\ \left|\begin{array}{cc}-3& 3\\ 2& 3\end{array}\right|& \left|\begin{array}{cc}2& 3\\ 2& 3\end{array}\right|& \left|\begin{array}{cc}2& -3\\ 2& 2\end{array}\right|\end{array}\right]=\left[\begin{array}{ccc}10& -5& -10\\ 0& -5& 5\\ -15& 0& 10\end{array}\right]$

$=\left[\begin{array}{ccc}(-1{)}^{1+1}(10)& (-1{)}^{1+2}(-5)& (-1{)}^{1+3}(-10)\\ (-1{)}^{2+1}(0)& (-1{)}^{2+2}(-5)& (-1{)}^{2+3}(5)\\ (-1{)}^{3+1}(-15)& (-1{)}^{3+2}(0)& (-1{)}^{3+3}(10)\end{array}\right]=\left[\begin{array}{ccc}10& 5& -10\\ 0& -5& -5\\ -15& 0& 10\end{array}\right]$

Transpose of the matrix A with co-factors

${{\mathrm{C}}_{\mathrm{A}}}^{\mathrm{T}}={\left[\begin{array}{ccc}10& 5& -10\\ 0& -5& -5\\ -15& 0& 10\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{ccc}10& 0& -15\\ 5& -5& 0\\ -10& -5& 10\end{array}\right]=\mathrm{adj}\mathrm{A}$

The inverse of the matrix $\mathrm{A},{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\text{ adj }\mathrm{A}$

$=\frac{1}{-25}\left[\begin{array}{ccc}10& 0& -15\\ 5& -5& 0\\ -10& -5& 10\end{array}\right]=\frac{1}{25}\left[\begin{array}{ccc}-10& 0& 15\\ -5& 5& 0\\ 10& 5& -10\end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc}-2& 0& 3\\ -1& 1& 0\\ 2& 1& -2\end{array}\right]$