Suppose two $n\times n$ matricies, $A$ and $B$, how many possible solutions are there. Suppose i construct a $n\times n$ matrix $C$, by multiplying two $n\times n$ matrices $A$ and $B$ i.e. $AB = C$. Given $B$ and $C$, how many other $A$'s can yield $C$ also i.e. is it just the exact $A$, infinitely many other $A$'s or no other $A$'s. There are no assumptions made about the invertability of $A$ and $B$. In the case that $A$ and $B$ are invertable there exists only one such $A$. 

In general, there could be infinitely many $A$.  
Given two solutions $A_1B=C$ and $A_2B=C$, we see that $(A_1-A_2)B=0$
So, if there is at least one solution to $AB=C$, you can see that there are as many solutions to $AB=C$ as there are to $A_0B=0$
Now if $B$ is invertible, the only $A_0$ is $A_0=0$.
If $A_0B=0$ then $(kA_0)B=0$ for any real $k$, so if there is a non-zero root to $A_0B=0$ there are infinitely many.
So you have to show that if $B$ is not invertible, then there is at least one matrix $A_0\neq 0$ such that $A_0B=0$.
Indeed, the set of such $A_0$ forms a vector space which depends on the "rank" of $B$.
