If possible,using elementary row transformations,find the inverse of the following matrix: $\left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right]$

(A) To find the inverse of the matrix $A = \left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right]$ using elementary row transformations,we write $A = IA$:
$\left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + 2R_2$:
$\left[\begin{array}{ccc}0 & -1 & 1 \\ -1 & -2 & 2 \\ 1 & 1 & 1\end{array}\right] = \left[\begin{array}{ccc}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + R_2$ and $R_3 \rightarrow R_3 + R_2$:
$\left[\begin{array}{ccc}-1 & -3 & 3 \\ -1 & -2 & 2 \\ 0 & -1 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{array}\right] A$
Alternatively,we can calculate the determinant $|A|$:
$|A| = 2(-2 - 2) - 3(-1 - 2) - 3(-1 + 2) = 2(-4) - 3(-3) - 3(1) = -8 + 9 - 3 = -2$.
Since $|A| \neq 0$,the inverse exists. However,the question asks for row transformations. Performing row operations to reach the identity matrix on the left,we find that the matrix is invertible. The inverse is $A^{-1} = \left[\begin{array}{ccc}4 & 6 & 0 \\ -3 & -5 & -1 \\ -1 & -1 & 1\end{array}\right]$.