Matrix inverse in mod6 Does the matrix $$\left[\begin{array}{cc} 5 & 2\\ 0 & 3\end{array}\right]$$
have a multiplicative inverse belonging to integer mod 6?
 A: Integer mod 6, or $\mathbb{Z}_6$, is not a field...and so some elements in it has no multiplicative inverse. The integer $3$ is one example: we need $x \in \{1,2,3,4,5\}$ such that \begin{equation} 3\cdot x = 1\mod 6,\end{equation} but as you can see such an $x$ does not exist. 
Now check the last row of the given matrix - for this matrix to have an inverse we must find a $2 \times 2$ matrix with entries from $\mathbb{Z}_6$ such that it's second column has the property: \begin{equation} \begin{bmatrix}0 & 3\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=1 (\mod 6). \end{equation} This means we must have $3 \cdot y =1 (\mod 6)$, which is not possible, as we have seen above.  
A: Instead of thinking about the matrix $\begin{bmatrix}1/5 & -2/5\\ 0& 1/3\end{bmatrix}$, you should think about the matrix $$\begin{bmatrix}1\cdot5^{-1} & -2\cdot 5^{-1}\\ 0& 1\cdot3^{-1}\end{bmatrix},$$ where the inverses are computed in the ring $\mathbb{Z}/\mathbb{Z}_6$ (or $\mathbb{Z}_6$, if you prefer that notation).
