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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 $e^{-At}$?

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  • $\begingroup$ Yes. (...and some more characters) $\endgroup$ Apr 17 '13 at 22:30
  • $\begingroup$ Does that identity rule also apply to a symmetric matrix? $\endgroup$
    – develarist
    Dec 11 '20 at 6:37
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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 $$ \left( e^A \right)^{-1} = e^{-A}. $$

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    $\begingroup$ 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? $\endgroup$
    – shardy
    Apr 17 '13 at 22:49
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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.

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  • $\begingroup$ I see. Thanks ! $\endgroup$
    – shardy
    Apr 17 '13 at 22:57

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