Prove that the product of two invertible matrices also invertible I am working on a homework problem, but I am lacking some understanding. 
Here is the problem:
Let $A$ and $B$ be invertible $n \times n$ matrices with $\det(A) = 3$ and $\det(B) = 4$.
I know that the product matrix of two invertible matrices must be invertible as well, but I am not sure how to prove that. I am trying to show it through the product of determinants if possible.
 A: Yes, since $\det(AB) = \det(A)\cdot \det(B) = 3\cdot 4 = 12 \neq 0$.
A: $C$ is invertible iff for all $y$ there is some $x$ such that $Cx=y$.
Suppose $A,B$ are invertible and choose some $y$. Then there is some $z$ such that $Az=y$, and there is some $x$ such that $Bx = z$. Hence $ABx=y$, and
so we see that $AB$ is invertible.
A: It is that $(AB)^{-1}=B^{-1}A^{-1}$, because $AB(AB)^{-1}=ABB^{-1}A^{-1}=1\!\!1$, but only for $n\times n$ matrices.
A: lets assume that C is a product of two invertible matrices .i.e. $C = AB$, there exists $A^{-1}$ such that $A^{-1}A = I = AA^{-1}$  and there exists $B^-1$ such that $B^{-1}B = I = B^{-1}B$. 
We need to prove that for C there exist a Right Inverse D such that $CD = I$ as well as a Left Inverse E such that $EC = I$.
Lets prove that there exists a Right Inverse :
we know, $C=AB$
Multiplying $B^{-1}$ from right on both sides,
$=> CB^{-1} = ABB^{-1} => CB^{-1} = AI => CB^{-1} = A $
Multiplying both sides by $A^{-1}$ from right,
$=> CB^{-1}A^{-1} = AA^{-1} => CB^{-1}A^{-1} = I $
This proves that the Right Inverse(D) for C is $B^{-1}A^{-1}$
Note: we could multiply $A^{-1}$ and $B^{-1}$ on both sides because they exist and along the way, we were just trying to reduce right hand side to Identity Matrix.
A similar proof can be given to prove that there exists a Left Inverse for C(this time multiplication has to be done from left side) and you will come to know that the Left Inverse is also the same.
