The inverse of \(AB\) is the product of \(A\) and \(B\) in the reverse order:
$$(AB)^{-1}=B^{-1}A^{-1}$$
Proof
Suppose that matrices \(A\) and \(B\) are invertible. Then we have
\begin{eqnarray*}(AB)(B^{-1}A^{-1})&=&A(BB^{-1})A^{-1}=AA^{-1}=I\\(B^{-1}A^{-1})(AB)&=&B^{-1}(A^{-1}A)B=B^{-1}B=I\end{eqnarray*}
where \(I\) is an identity matrix.
This shows that the inverse matrix of \(AB\) is \(B^{-1}A^{-1}\):
$$(AB)^{-1}=B^{-1}A^{-1}$$