Let A be a $$3 \times 3$$ matrix such that
$$A\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$
Then A$$^{-1}$$ is:

We are given the matrix equation:

$$A \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

Let us denote the matrix on the left as $$ B $$, so:

$$B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}$$

And the matrix on the right as $$ C $$, so:

$$C = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

The equation becomes $$ A B = C $$. To find $$ A^{-1} $$, we rearrange the equation. Multiplying both sides on the right by $$ B^{-1} $$ gives $$ A = C B^{-1} $$. Then, taking the inverse of both sides:

$$A^{-1} = (C B^{-1})^{-1} = B C^{-1}$$

So, we need to compute $$ B C^{-1} $$. First, we find $$ C^{-1} $$. Notice that $$ C $$ is a permutation matrix that cycles the rows: applying $$ C $$ to a vector $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$ gives $$ \begin{bmatrix} z \\ x \\ y \end{bmatrix} $$. The inverse permutation should map $$ \begin{bmatrix} z \\ x \\ y \end{bmatrix} $$ back to $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$, which corresponds to the matrix:

$$C^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

Verification:

$$C^{-1} \begin{bmatrix} z \\ x \\ y \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} z \\ x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

Now, we compute $$ B C^{-1} $$:

$$B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}, \quad C^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$

Perform matrix multiplication:

Element at row 1, column 1: $$ (1)(0) + (2)(0) + (3)(1) = 0 + 0 + 3 = 3 $$

Element at row 1, column 2: $$ (1)(1) + (2)(0) + (3)(0) = 1 + 0 + 0 = 1 $$

Element at row 1, column 3: $$ (1)(0) + (2)(1) + (3)(0) = 0 + 2 + 0 = 2 $$

Element at row 2, column 1: $$ (0)(0) + (2)(0) + (3)(1) = 0 + 0 + 3 = 3 $$

Element at row 2, column 2: $$ (0)(1) + (2)(0) + (3)(0) = 0 + 0 + 0 = 0 $$

Element at row 2, column 3: $$ (0)(0) + (2)(1) + (3)(0) = 0 + 2 + 0 = 2 $$

Element at row 3, column 1: $$ (0)(0) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1 $$

Element at row 3, column 2: $$ (0)(1) + (1)(0) + (1)(0) = 0 + 0 + 0 = 0 $$

Element at row 3, column 3: $$ (0)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1 $$

So, the product is:

$$B C^{-1} = \begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix}$$

Therefore, $$ A^{-1} = \begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix} $$.

Comparing with the options:

Option A: $$ \begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix} $$

Option B: $$ \begin{bmatrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{bmatrix} $$

Option C: $$ \begin{bmatrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix} $$

Option D: $$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{bmatrix} $$

Our result matches Option A.

Hence, the correct answer is Option A.