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

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 $e^{-At}$?

• Yes. (...and some more characters) Apr 17 '13 at 22:30
• Does that identity rule also apply to a symmetric matrix? Dec 11 '20 at 6:37

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 $$\left( e^A \right)^{-1} = e^{-A}.$$
• I understand but how does that imply $e^{-At} = \frac{1}{4} \begin{bmatrix} -e^{t} + 5e^{-3t} & e^{t} - e^{-3t} \\ -5e^{t} + 5e^{-3t} & 5e^{t} - e^{-3t} \end{bmatrix}$? Can i just switch all $e^t$ with $e^{-t}$ in $e^{At}$ to get $e^{-At}$? if yes, why? Apr 17 '13 at 22:49
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.