Find the inverse matrix of the following matrix, $$A=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}$$ over $F_3$, i.e., all elements are one of $\{0, 1, 2\}$.
I first found its determinant. (I found it using first row as coefficients.)
\begin{align} \det A &= 0 \cdot \begin{bmatrix}0&1\\1&0\end{bmatrix} - 1 \cdot \begin{bmatrix}1&1\\1&0\end{bmatrix} + 1 \cdot \begin{bmatrix}1&0\\1&1\end{bmatrix}\\ &= 1 + 1 \\ &= 2. \end{align}
Then, the inverse of $A$ is
\begin{align} A^{-1} &= \frac{1}{\det A} \begin{bmatrix} +\left|\begin{array}{cc}0&1\\1&0\end{array}\right| & -\left|\begin{array}{cc}1&1\\1&0\end{array}\right| & +\left|\begin{array}{cc}1&0\\1&1\end{array}\right|\\ -\left|\begin{array}{cc}1&1\\1&0\end{array}\right| & +\left|\begin{array}{cc}0&1\\1&0\end{array}\right| & -\left|\begin{array}{cc}0&1\\1&1\end{array}\right|\\ +\left|\begin{array}{cc}1&1\\0&1\end{array}\right| & -\left|\begin{array}{cc}0&1\\1&1\end{array}\right| & +\left|\begin{array}{cc}0&1\\1&0\end{array}\right| \end{bmatrix}\\ &= \frac12 \begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}\\ &= \begin{bmatrix}1&0.5&0.5\\0.5&1&0.5\\0.5&0.5&1\end{bmatrix}, \end{align} where $\left|M\right|$ represents determinent of M.
The problem is that $0.5$ is not a member of $F_3$.
In this case, is the answer $!\exists ~ A^{-1}$ ? or is my solution process wrong?
Thank you for reading my question.