There is an easier way to compute manually $e^{tA}$? I have this matrix
$$A:=\begin{bmatrix}0&-1&1\\0&0&1\\-1&0&1\end{bmatrix}$$
And I have tested that $A^3\neq I$ but $A^4=I$, and I want to find $e^{tA}$. Then what I did was
$$e^{tA}=\sum_{k=0}^\infty\frac{(tA)^k}{k!}=I\sum_{k=0}^\infty\frac{t^{4k}}{(4k)!}+A\sum_{k=0}^\infty\frac{t^{4k+1}}{(4k+1)!}+A^2\sum_{k=0}^\infty\frac{t^{4k+2}}{(4k+2)!}+A^3\sum_{k=0}^\infty\frac{t^{4k+3}}{(4k+3)!}$$
but I dont know if I can find $e^{tA}$ in an easier and computable-by-hand way than the above. 
The above seems painful to do it manually. So, if someone have some idea to improve the manual computation of $e^{tA}$ I will like to know it.

Let$$P=\begin{pmatrix}1-i&1+i&0\\-i&i&1\\1&1&1\end{pmatrix}.$$The columns of $P$ are eigenvectors of $A$. Then$$P^{-1}=\frac14\begin{pmatrix}1+i & -1+i & 1-i \\ 1-i & -1-i & 1+i \\-2 & 2 & 2\end{pmatrix}$$and$$P^{-1}.A.P=\begin{pmatrix}i&0&0\\0&-i&0\\0&0&1\end{pmatrix}.$$Therefore$$P^{-1}.e^{tA}.P=\begin{pmatrix}e^{ti}&0&0\\0&e^{-ti}&0\\0&0&e^t\end{pmatrix},$$which means that $e^{tA}$ is equal to$$\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix},$$where\begin{align}a_{11}&=\cos (t)\\a_{12}&=-\sin (t)\\a_{13} &= \sin (t) \\ a_{21}&=
 \frac{1}{2} \left(\cos (t)-\cosh (t)+\sin (t)-\sinh (t)\right)\\a_{22} &= \frac{1}{2} \left(\cos(t)+\cosh (t)-\sin (t)+\sinh (t)\right)\\a_{23} &= \frac{1}{2} \left(-\cos (t)+\cosh (t)+\sin (t)+\sinh (t)\right) \\a_{31}&= \frac{1}{2} \left(\cos (t)-\cosh (t)-\sin (t)-\sinh (t)\right)\\ a_{32}&= \frac{1}{2} \left(-\cos (t)+\cosh (t)-\sin (t)+\sinh (t)\right)\\a_{33} &= \frac{1}{2} \left(\cos (t)+\cosh (t)+\sin (t)+\sinh (t)\right).\end{align}
