Failure during calculating the matrix exponential, but where? I have to calculate $e^{At}$ of the matrix $A$. We are learned to first compute $A^k$, by just computing $A$ for a few values of $k$, $k=\{0\ldots 4\}$, and then find a repetition. $A$ is defined as follows:
$$ A =
\begin{bmatrix} 
-2 & 2& 0 \\ 
0 & 1 & 0 \\ 
1 & -1 & 0 
\end{bmatrix}
$$
Because I couldn't find any repetition I used Wolfram|Alpha which gave me the following, http://goo.gl/JxyIg:
$$
\frac{1}{6}
\begin{bmatrix}
3(-1)^k2^{k+1} & 2(2-(-1)^k2^{k+1}) & 0 \\
0 & 6 & 0 \\
3(-(-2)^k+0^k) & 2(-1+(-2)^k) & 6*0^k
\end{bmatrix}
$$
Then $e^{At}$ is calculated as followed (note that $\sum_{k=0}^{\infty}\frac{0^kt^k}{k!} = e^{0t} = 1$, using that $0^0 = 1$):
$$ e^{At} = 
\begin{bmatrix}
\frac{1}{6}\sum_{k=0}^{\infty}\frac{3(-1)^k2^{k+1}t^k}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{2(2-(-1)^k2^{k+1})t^k}{k!} & 0 \\
0 & \frac{1}{6}\sum_{k=0}^{\infty}\frac{6t^k}{k!} & 0 \\
\frac{1}{6}\sum_{k=0}^{\infty}\frac{3(-(-2)^k+0^k)t^k}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{2(-1+(-2)^k)}{k!} & \frac{1}{6}\sum_{k=0}^{\infty}\frac{6^k*0^k}{k!}
\end{bmatrix}
$$
Now this matrix should give as a answer
$$
\begin{bmatrix} 
e^{-2t} & e^{2t} & 0 \\ 
0 & e^{t} & 0 \\ 
e^{t} & e^{-t} & 1
\end{bmatrix}
$$
Now when I compute this answer of $e^{At}$, I get different answers for some elements. Only the elements $A_{11} = e^{-2t}$, $A_{13} = A_{21} = A_{23} = A_{33} = 1$ and $A_{22} = e^t$. However when I calculate $A_{12}$ I get the following:
$$
A_{12}=\frac{1}{6}\sum_{k=0}^{\infty}\frac{2(2-(-1)^k2^{k+1})t^k}{k!}=\frac{2}{6}\left(\sum_{k=0}^{\infty}\frac{2t^k}{k!}-\sum_{k=0}^{\infty}\frac{(-1)^k2^{k+1}t^k}{k!}\right)=\frac{4}{6}\left(\sum_{k=0}^{\infty}\frac{t^k}{k!}-\sum_{k=0}^{\infty}\frac{(-1)^k2^{k}t^k}{k!}\right)=\frac{4}{6}\left(e^t-e^{-2t}\right)
$$
Which is of course not equal to $e^{2t}$. Where do I make a mistake? Or does maybe Wolfram|Alpha make a mistake, I know it is correct for $0\ldots 4$.
 A: Corrected : Your $A^k$ is right but for $e^{At}$ you should just have multiplied every term by $\frac {t^k}{k!}$ before computing the sum. So no $6^k$ in the  central term for example.
Further (as explained by Robert Israel) for $k=0$ your $A^k$ expression is still valid with only diagonal $1$ terms (so no $1$ elsewhere). 
Last you seem to be supposing that $e^{At}$ will simply be the exponential of each term : this is not true as shown by your $A_{1,2}$ term (i.e. the 'should give...' part is not right). I'll add that, in Mathematica, you must use MatrixExp to compute a matrix exponential and not Exp that returns the exponential of the individual terms! Result :

Hoping this helped more,
A: In general, you would be better off by finding a diagonalization of $A$ and exponentiating the diagonal matrix, i.e.
$e^{At} = e^{VDV^Tt} = \sum_k \frac{(VDV^Tt)^k}{k!} = \sum_k V\frac{D^k t^k}{k!} V^T = Ve^{Dt}V^T$,
(since $V$ is orthonormal). Then, $e^D$ simply exponentiates the diagonal of $D$, premultiplying every diagonal entry with $t$, and you have to multiply things out...
