Find the inverse of the matrix $A = \left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$,if it exists.

Let $A = \left[\begin{array}{rrr}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$.
First,we find the determinant of $A$ $(|A|)$:
$|A| = 2(4 - (-6)) - (-3)(4 - 9) + 3(-4 - 6)$
$|A| = 2(10) + 3(-5) + 3(-10)$
$|A| = 20 - 15 - 30 = -25$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix $C_{ij}$:
$C_{11} = +(4+6) = 10, C_{12} = -(4-9) = 5, C_{13} = +(-4-6) = -10$
$C_{21} = -(-6+6) = 0, C_{22} = +(4-9) = -5, C_{23} = -(-4+9) = -5$
$C_{31} = +(-9-6) = -15, C_{32} = -(6-6) = 0, C_{33} = +(4+6) = 10$
The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:
$adj(A) = \left[\begin{array}{rrr}10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10\end{array}\right]$
Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-25} \left[\begin{array}{rrr}10 & 0 & -15 \\ 5 & -5 & 0 \\ -10 & -5 & 10\end{array}\right] = \left[\begin{array}{rrr}-2/5 & 0 & 3/5 \\ -1/5 & 1/5 & 0 \\ 2/5 & 1/5 & -2/5\end{array}\right]$.