Q:

|$A$|$=0$ implies |$AB$|$=0$

Without using the fact that $\det AB=\det A\det B$ show that if:
$|A|=0$ and $A,B$ are $n\times n$ matrices then $|AB|=0$
So for contradiction assume that |$AB|\neq 0$ then there exists an $n\times n$ matrix $C$ such that:
$ABC=I_n$ and $CAB=I_n$.
So all I need to show now is $BCA=I_n$ and then I have an $n\times n$ matrix $BC$ such that $A^{-1}=BC$, which is a contradiction.
How do I show $BCA=I_n$?

A:

Hint:
$$
(AB)C=I \Rightarrow A(BC)=I \Rightarrow BC=A^{-1}
$$

