If $A,B$ are invertible so $AB$ is invertible I thought about the following proofs but I am not sure about them 1. there is $C,D$ so that $AC=CA=I$ and $BD=DB=I \rightarrow CABD=I \rightarrow$ due to associativity roles is no matrix $E$ so that $EAB=ABE=I$ 2. let look at $A=\begin{pmatrix} 1 & 0\\\ 3 & 3 \end{pmatrix}$, $B=\begin{pmatrix} 0 & 3\\\ 2 & -3 \end{pmatrix}$, $AB=\begin{pmatrix} 0 & 3\\\ 6 & 0 \end{pmatrix}$ so $A,B$ is invertible but $AB$ is not, but using Matlab $AB$ does have an inverse.

The product of two invertible matrices is always invertible. If $A^{-1}$ and $B^{-1}$ both exist, then $(AB)^{-1} = B^{-1}A^{-1}$.

This can be verified by pre and post multiplication.

Premultiplication:

$B^{-1}A^{-1}AB = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I $

The postmultiplication should be obvious.

The matrix product in your example is indeed invertible as it has a nonzero determinant.