# 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.

• This really depends on what facts you have to work with. It turns out there are a lot of equivalent ways to say a matrix is invertible, but you may not have seen some of those in class yet and hence can't really use them. For instance, one of the answers here used the fact that $A$ is invertible iff $\det(A)\neq0$. Have you seen this before? You could also explicitly find an inverse, since you know $A$ and $B$ both have inverses. – Alex Mathers Apr 28 '16 at 6:16
• Thanks Alex! I know how to prove it using the fact that A is invertible, but I am trying to understand how would the prove look like using the determinants since otherwise I do not see why would I need them in the problem. – Romaldowoho Apr 28 '16 at 6:20

Yes, since $\det(AB) = \det(A)\cdot \det(B) = 3\cdot 4 = 12 \neq 0$.
$C$ 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.
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.