Question: Where can I exponentiate a $3\times 3$ matrix like $\left[\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{array}\right]$
 online?

Is there some website where this is possible?

Thank you very much.

Answer: 
Depending on which matrix exponential you want, you can use:

* [*Wolfram Alpha 1st option: $e^{A}$*](http://www.wolframalpha.com/input/?i=matrixexp%28%7B%7B0%2C1%2C0%7D%2C%7B1%2C0%2C0%7D%2C%7B0%2C0%2C0%7D%7D%29)
* [*Wolfram Alpha 2nd option: $e^{At}$*](http://www.wolframalpha.com/input/?i=matrixexp%28t+%7B%7B0%2C1%2C0%7D%2C%7B1%2C0%2C0%7D%2C%7B0%2C0%2C0%7D%7D%29)

This is actually a command in Mathematica.

I would be surprised if were not available in other [*CAS programs*](http://en.wikipedia.org/wiki/List_of_computer_algebra_systems) and some of those are online, like [*Sage*](http://www.sagemath.org/doc/reference/matrices/sage/matrix/matrix_symbolic_dense.html), [*Maxima*](http://www.ma.utexas.edu/pipermail/maxima/2006/003031.html) and others.
Somewhat perversely, could I suggest you first try analytically, by [Sylvester's formula](https://en.wikipedia.org/wiki/Sylvester%27s_formula), or, more specifically, [simpler methods as below](https://en.wikipedia.org/wiki/Matrix_exponential#Computing_the_matrix_exponential)? 

Still, in your specific case, this would be overkill: your matrix is trivially diagonalizable,
$$M=\left[\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{array}\right]=\frac{1}{\sqrt2}\left[\begin{array}{ccc}
1 & 1 & 0\\
1 & -1 & 0\\
0 & 0 & {\sqrt2}
\end{array}\right] ~~
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 0
\end{array}\right]~\frac{1}{\sqrt2}\left[\begin{array}{ccc}
1 & 1 & 0\\
1 & -1 & 0\\
0 & 0 & {\sqrt2}
\end{array}\right]\equiv R^{-1} D R,$$
for the obviously defined diagonal matrix *D* and diagonalizing rotations *R*.

It then immediately follows that 
$$
\exp M= R^{-1} e^D R= \left[\begin{array}{ccc}
\cosh 1 & \sinh 1 & 0\\
\sinh 1 & \cosh 1 & 0\\
0 & 0 & 1
\end{array}\right].
$$

Sylvester's formula generalizes to non-diagonalizable matrices, but my sense is you are dealing with diagonalizable ones.