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 $0\ldots 4$.