If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]_{3 \times 3}$, then $A^{-1}=$

The determinant of $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ evaluates to $\det(A) = 3[(-3)(1) - (4)(-1)] + 3[(2)(1) - (4)(0)] + 4[(2)(-1) - (-3)(0)] = 3(1) + 3(2) + 4(-2) = 1$.

The cofactor matrix is computed as $C = \begin{bmatrix} 1 & -2 & -2 \\ -1 & 3 & 3 \\ 0 & -4 & -3 \end{bmatrix}$, yielding the adjoint $\operatorname{adj}(A) = C^\top = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

Thus, $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

Computing $A^2 = A \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$ and then $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$ shows that $A^{-1} = A^3$.

The result corresponds to option C.