I was just wondering whether my solution is correct or not and if it isn't, where I went wrong?
Find $ e^A $ where $ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $
So what I did is I saw that
$$ e^A = \sum_{n=0}^{\infty} \frac{\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}^{n}}{n!}$$
and I figured out that when n is even, the resulting matrix is the identity matrix and when n is odd it just turns out to be A itself, so now I transformed it into
$$ e^A = \sum_{n=0}^{\infty} \frac{\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}}{2n!} +\sum_{n=0}^{\infty} \frac{\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}}{(2n+1)!} $$ after this I don't really know what to do, they look similar to the power series of cos and sin though.
$A$ is an order-$2$ permutation matrix, hence:
$$ e^{A}=\sum_{n\geq 0}\frac{A^n}{n!} = A\sum_{n\geq 0}\frac{I}{(2n+1)!}+I\sum_{n\geq 0}\frac{I}{(2n)!} = \sinh(1) A + \cosh(1) I = \begin{pmatrix}\cosh(1)&\sinh(1)\\\sinh(1)&\cosh(1)\end{pmatrix}=\frac{1}{2e}\begin{pmatrix}e^2+1&e^2-1\\e^2-1&e^2+1\end{pmatrix}. $$