When dealing with matrix exponentials, can we assume $e^{-A}=(e^A)^{-1}$?

Question

Of course, $e^{-x}=(e^x)^{-1}$, but in this case we have a matrix $A$. I think it is in fact true that $e^{-A}=(e^A)^{-1}$, but how would I go about proving this?

Answer 1

Because\begin{align}e^Ae^{-A}&=e^{A+(-A)}\text{ (since $A$ and $-A$ commute)}\\&=e^0\\&=\operatorname{Id}.\end{align}