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.

The usual way to compute analytic functions of matrices is to exploit the fact they behave well under conjugation:
$$\begin{align}
 f(PAP^{-1}) &= \sum_n a_n (PAP^{-1})^n
\\&= \sum_n a_n P A^n P^{-1}
\\&= P \left( \sum_n a_n A^n \right) P^{-1} 
\\&= P f(A) P^{-1} \end{align}$$
and thus
$$ f(A) = P^{-1} f(PAP^{-1}) P $$
The plan is to arrange for $B=PAP^{-1}$ to be a matrix for which $f(B)$ is easy to compute. Diagonal matrices, in particular, are usually the simplest:
$$\begin{align}
 f\left( \begin{pmatrix}b_1 & 0 & \ldots
\\ 0 & b_2 & \ldots
\\ \vdots & \vdots & \ddots \end{pmatrix} \right)
&= \sum_n a_n \begin{pmatrix}b_1 & 0 & \ldots
\\ 0 & b_2 & \ldots
\\ \vdots & \vdots & \ddots \end{pmatrix}^n
\\&=\sum_n a_n  \begin{pmatrix}b_1^n & 0 & \ldots
\\ 0 & b_2^n & \ldots
\\ \vdots & \vdots & \ddots \end{pmatrix}
\\&= \begin{pmatrix}\sum_n a_n b_1^n & 0 & \ldots
\\ 0 & \sum_n a_n b_2^n & \ldots
\\ \vdots & \vdots & \ddots \end{pmatrix}
\\&= \begin{pmatrix}f(b_1) & 0 & \ldots
\\ 0 &  f(b_2) & \ldots
\\ \vdots & \vdots & \ddots \end{pmatrix}
\end{align}
$$
So if you can diagonalize $A$, this above plan will let you compute any analytic function of $A$.
