Given a solution to find a matrix For $e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t}  & e^{2t}+e^{-t}\end{bmatrix}$ for all t $\in$ $\mathbb{R}$. how to find A?
 A: The matrix exponential is given by:
$$\tag 1 e^{At} = \sum_{k=0}^{n-1} \alpha_k A^k$$
where the $\alpha_i$'s  are determined from the set of equations given by the eigenvalues of A, as:
$$\tag 2 e^{\lambda_i t} = \sum_{k=0}^{n-1} \alpha_k \lambda_i^k$$
We are given:
$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t}  & e^{2t}+e^{-t}\end{bmatrix}$$
From this, we know we have a characteristic polynomial of:
$$\lambda^2-\lambda -2 = 0 \implies \lambda_1 = -1, ~\lambda_2 = 2$$
From $(1)$, we have $n = 2$ (the number of eigenvalues) and can write:
$$\tag 3 e^{At} = \alpha_0 ~I + \alpha_1 ~ A$$
From $(2)$, we can write:
$$\alpha_0 - \alpha_1 = e^{-t} \\ \alpha_0 + 2 \alpha_1 = e^{2t}$$
Solving this $~2x2~$ yields:
$$\alpha_0 = \dfrac{1}{3} e^{2t} + \dfrac{2}{3}e^{-t}, ~~ \alpha_1 = \dfrac{1}{3}e^{2t} - \dfrac{1}{3}e^{-t}$$
From $(3)$, we have:
$$A = \dfrac{e^{At}-\alpha_0 I}{\alpha_1}$$
Substituting all of the information, we arrive at:
$$A = \dfrac{1}{2}\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}$$
Of course, we can use this calculated $A$ and determine $e^{At}$, yielding:
$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t}  & e^{2t}+e^{-t}\end{bmatrix}$$
A: $\newcommand{\+}{^{\dagger}}
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$$
\left.\mbox{Note that}\ A = \totald{\pars{\expo{At}}}{t}\right\vert_{t\ =\ 0}
$$

\begin{align}
\color{#00f}{\large A} &=
\totald{}{t}\bracks{\half\pars{\begin{array}{cc}
\expo{2t} + \expo{-t} & \expo{2t} - \expo{-t}
\\[2mm]
\expo{2t} - \expo{-t}  & \expo{2t} + \expo{-t}
\end{array}}}_{t\ =\ 0}
=
\half\pars{\begin{array}{cc}
2\expo{2t} - \expo{-t} & 2\expo{2t} + \expo{-t}
\\[2mm]
2\expo{2t} + \expo{-t}  & 2\expo{2t} - \expo{-t}
\end{array}}_{t\ =\ 0}
\\[3mm]&=
\color{#00f}{\large\half\pars{\begin{array}{cc}
1 & 3
\\[2mm]
3  & 1
\end{array}}}
\end{align}

