When calculating the inverse of a normal matrix, for $\displaystyle M = \begin{pmatrix}

x & y \\

z & v

\end{pmatrix}$ for example, you would simply use $\displaystyle \frac{1}{xv-yz}\begin{pmatrix}

v & -y \\

-z & x

\end{pmatrix}$.

However, if your matrix was $\displaystyle \begin{pmatrix}

\bar{1} & \bar{2} \\

\bar{2} & \bar{0}

\end{pmatrix}$ in $\displaystyle \mathbb{Z}_3$, the inverse would be:

$\displaystyle (-\bar{4})^{-1}\begin{pmatrix}

\bar{0} & \bar{-2} \\

\bar{-2} & \bar{1}

\end{pmatrix} = (\bar{2})^{-1}\begin{pmatrix}

\bar{0} & \bar{1} \\

\bar{1} & \bar{1}

\end{pmatrix}$.

Ok up to here, just forgotten how to calculate $\displaystyle (\bar{2})^{-1}$

Thanks in advance