Inverse of a matrix exponential, ${(e^{At})}^{-1} = {e^{-At}}$

Question

Consider the matrix exponential $$ e^{At} = \frac{1}{4} \begin{bmatrix} -e^{-t} + 5e^{3t} & e^{-t} - e^{3t} \\ -5e^{-t} + 5e^{3t} & 5e^{-t} - e^{3t} \end{bmatrix} $$

And $$ {(e^{At})}^{-1} = {e^{-At}} $$

What does that identity mean? Can I just multiply the exponent by $-1$ to find $e^{-At}$?

Answer 1

If two square matrices $A$ and $B$ commute ($AB = BA$), then $$ e^A e^B = e^{A+B}. $$

Now, $A$ and $-A$ always commute, so $$ e^A e^{-A} = e^{A+(-A)} = e^{0} = I. $$

This shows that $$ \left( e^A \right)^{-1} = e^{-A}. $$

Answer 2

If you know $e^{At}$, then to get $e^{-At} = e^{A(-t)}$, you just replace $t$ with $-t$ in the formula for $e^{At}$. The other answer is correct, but this answers the question you posed in your comment.