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 $A$. 

Two cases: either $B$ is invertible, and is this case $A$ is uniquely determined, or $B$ is not invertible. We can find two invertible matrices $P$ and $Q$ such that $P\operatorname{diag}(1,\ldots,1,0,\ldots,0)Q=B$ (there is at least one zero). Let $D:=\operatorname{diag}(1,\ldots,1,0,\ldots,0)$. The problem is to find $A$ such that $APD=CQ^{-1}$. Since when we multiply $AP$ by $D$ the terms one the last column of $AP$ doesn't matter, there are infinitely many $AP$ which work and since $P$ is invertible, infinitely many $A$.
