|$A$|$=0$ 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:
$$
(AB)C=I \Rightarrow A(BC)=I \Rightarrow BC=A^{-1}
$$
A: It is a general result about square matrices, that if $AB=I$ ($A$ and $B$ are square matrices), then also $BA=I$.
A: Assume that $\det (B) \neq 0$, then we have $\operatorname{range}(B) = \mathbb{R}^n$. From the fact that $\det(A) =0$, we know that there exist a $\mathbf{x} \in \mathbb{R}^n\setminus\{ \mathbf{0}\}$ such that $A\mathbf{x}=\mathbf{0} \in \mathbb{R}^n$. Let $\mathbf{y} \in \mathbb{R}^n\setminus\{ \mathbf{0}\}$ such that $B\mathbf{y}=\mathbf{x}$. Then we have
$$AB\mathbf{y} = A \mathbf{x}=\mathbf{0}$$
Hence $\ker (AB) \neq \{ \mathbf{0}\}$ which implies that $\det (AB) =0$
The case with $\det (B) =0$ is even easier.
A: If $\det A=0$, then the rows of $A$ are linearly dependent:
$\lambda_1a_1+\ldots+\lambda_n a_n=0.$ In this case, $a_1B,\ldots,a_nB$ are the rows of $AB$, and we get $\lambda_1a_1B+\ldots+\lambda_na_nB=0.$
A: As you said $|AB|\neq 0\rightarrow $ there is a $C$ such that $CAB=ABC=I$
Lets look at $Rank(AB)=n$ and we know that $AB\leq min(Rank(A),Rank(B))$ so we have $n\leq min(Rank(A),Rank(B))$ which implies that Rank(A)=Rank(B)=n (Because both matrices are of $Rank=n$ and at least of $Rank=n$)  therefore $|A|\neq 0$
