Given matrix $e^{tA}$, find $A$ I tried answering the following question in an old exam:

Given matrix:

$$ e^{tA} = \begin{bmatrix}\frac{1}{2}(e^t+e^{-t}) &amp; 0 &amp; \frac{1}{2}(e^t-e^{-t})\\0 &amp; e^t &amp; 0\\\frac{1}{2}(e^t-e^{-t}) &amp; 0 &amp; \frac{1}{2}(e^t+e^{-t})\end{bmatrix}$$

Calculate A.

Where $$e^{tA} = \sum_{k=0}^{\infty} \frac{t^k}{k!}A^k$$

Hope I can get some help with this, as I don't think I can find the right way to approach it myself.
$$A=\left.\frac d{dt}e^{tA}\right|_{t=0}=\begin{bmatrix}0&amp;0&amp;1\\0&amp;1&amp;0\\1&amp;0&amp;0\end{bmatrix}.$$
The result given by José is fine if you know that the matrix  is for sure obtainable as $\exp(t A)$.

If you are doubtful (maybe it is a trick questions), you can use the alternative
$$A = e^{-t A}\left( \frac{d}{dt} e^{t A}\right) = \begin{pmatrix} 0&amp; 0 &amp; 1\\
0 &amp; 1 &amp;0\\
1 &amp;0 &amp;0 \end{pmatrix} $$
where you can check explicitly that $t$ drops out.