Find all nonzero matrices $A\in M_2(\Bbb Z_3)$ which are not invertible. Find all nonzero (with every element $\neq$0) matrices $A\in M_2(\Bbb Z_3)$ which are not invertible, and explain why they aren't invertible. 
It seems simple indeed, but I'm not sure how to solve this :s
Thanks for any help!!! 
 A: Since every element of our matrix is supposed to be non-zero, we have very few cases to consider. For at every position of the matrix, we have $2$ choices for the entry at that position, for a total of $2^4$. We can just go through these $16$ possibilities, and identify the matrices that are non-invertible. The two procedures we describe below are somewhat faster. 
We can use the criterion that a matrix is non-invertible if and only if the two rows are linearly dependent. Or else we can use the fact that a matrix over a field is invertible if and only if its determinant is non-zero.  
Solution $1$. The first row can be any of $1\quad 1$, or $1\quad 2$, or $2\quad 1$, or $2\quad 2$.
For each of these, there are precisely two second rows such that the first row and the second are linearly dependent. For example, if the first row is $1\quad 2$, the two second rows are $1$ times the first row, or $2$ times the first row, that is, $1\quad 2$ and $2\quad 1$.
Solution $2.$ Let $a$ and $d$ be the entries on the main diagonal. There are $4$ possibilities for the pair $(a,d)$. Our matrix will be non-invertible precisely if its determinant is $0$. This is the case if and only if $cb\equiv ad\pmod{3}$, where $b$ is the second entry in the top row, and $c$ is the first entry in the second row. 
Now we can do a case by case analysis. For example, if $a=1$ and $d=2$, we want $cb\equiv 2\pmod{3}$, giving $b=2$ and $c=1$ or the other way around. 
Remark: For the same problem with $\mathbb{Z}_p$ instead of $\mathbb{Z}_3$, the same ideas work. For example, let $p=13$. For any of the $(12)(12)$ first rows, there are $12$ second rows that give non-invertibility. 
A: Hint: $A \in M_2(\mathbb{Z}_3)$ is not invertible iff its rows $A_1$ and $A_2$ are colinear. In particular, you can deduce that there are $33$ non invertible matrices in $M_2(\mathbb{Z}_3)$.
