Given the matrix $A∈M_3(Z_9)$
$$ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 0 \\ 0 & 3 & 6 \\ \end{matrix} $$
The determinant is:
$\det(A)=18+18-24=3$
So the matrix is invertible because the determinant is different from zero.
The transposed matrix is:
$$ \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 3 \\ 3 & 0 & 6 \\ \end{matrix} $$
The matrix $A'$ is:
$$ \begin{matrix} 0 & 6 & 0 \\ 6 & 6 & 6 \\ 6 & 6 & 8 \\ \end{matrix} $$
So the inverse matrix is:
$$ \begin{matrix} 0 & 2 & 0 \\ 2 & 2 & 2 \\ 2 & 2 & 8/3 \\ \end{matrix} $$
Please tell me if my attempt is correct.
In fact, the non-existence of the inverse can be proven by row echelon form calculation. This is a more basic approach.
\begin{align} & \left[\begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0\\ 2 & 3 & 0 & 0 & 1 & 0\\ 0 & 3 & 6 & 0 & 0 & 1 \end{array}\right] \\ &\sim \left[\begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -1 & -6 & -2 & 1 & 0 \\ 0 & 3 & 6 & 0 & 0 & 1 \end{array}\right] (R_2 \leftarrow R_2 - 2R_1) \\ &\sim \left[\begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & 2 & -1 & 0 \\ 0 & 3 & 6 & 0 & 0 & 1 \end{array}\right] (R_2 \leftarrow -R_2) \\ &\sim \left[\begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & 2 & -1 & 0 \\ 0 & 0 & -12 & -6 & 3 & 1 \end{array}\right] (R_3 \leftarrow R_3 - 3R_2) \\ &= \left[\begin{array}{rrr|rrr} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & 2 & 8 & 0 \\ 0 & 0 & 6 & 3 & 3 & 1 \end{array}\right] \end{align}
One is tempted to divide the last row by $6$, but since we are in $\Bbb Z_9$, this is not permitted. In fact, the row operations above show that $A$ has no inverse in $M_3(\Bbb Z_9)$: If we omit the fourth and the fifth column of of the above augmented matrices, we are actually finding a solution to
$$\begin{bmatrix} 1&2&3\\ 2&3&0\\ 0&3&6 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}.$$ Through the above row echelon form, we deduce that $6x_3 = 1$ in $\Bbb Z_9$, which is absurd. Therefore, $A$ is noninvertible in $M_3(\Bbb{Z}_9)$.