How to find the sum of elements of $e^M$ where $M$ is a matrix. The question is 

If $M=\left(\begin{matrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{matrix}\right)$ and $e^M=I+M+\frac{1}{2!}M^2+\cdots$. If $e^M=[b_{ij}]$ then what is the value of $$\frac{1}{e}\sum_{i=1}^{3}\sum_{j=1}^{3}b_{ij}$$

I don't have any idea how to solve this but I tried to calculate the powers of $M$, and summing the series $e^M$ but that didn't give me a clear idea. How can I do this? Any help is appreciated. 
 A: Hint. Let $N=\pmatrix{0&1&0\\ 0&0&1\\ 0&0&0}$. Then $M=I+N$ and $N^3=0$. Therefore the power series for $e^N$ is a finite series and you can calculate $e^N$ easily. Now, using the fact that $e^{A+B}=e^Ae^B$ for commuting matrices $A,B$, you may calculate $e^M=e^{I+N}$ explicitly. The required quantity $\frac1e\sum_{i,j}b_{ij}$ is just the sum of all entries of $e^M$ divided by $e$.
A: You can prove by induction that
$$M^k = \left( \begin{matrix} 1 & k & \binom{k}{2} \\ 0 & 1 & k \\ 0&0&1\end{matrix} \right)$$
so that $$b_{11} =b_{22} =b_{33} = \sum_{k=0}^{\infty} \frac{1}{k!}=e$$
$$b_{12}=b_{23} = \sum_{k=0}^{\infty} \frac{1}{k!}k = e$$
$$b_{13} = \sum_{k=0}^{\infty} \frac{1}{k!}\frac{k(k-1)}{2}=e/2$$
i.e.
$$e^M = \left( \begin{matrix} e & e & e/2 \\ 0 & e & e \\ 0&0&e\end{matrix} \right)$$
can you conclude now?
A: you have the next equality $M^{k}=\left( \begin{array}{ccc}
1 & k & \frac{(k-1)k}{2} \\
0 & 1 & k \\
0 & 0 & 1 \end{array} \right)$
then you have the next result:
$$b_{11}=b_{22}=b_{33}=\sum_{k=0}^{\infty}\frac{1}{k!}=e$$
$$b_{12}=b_{23}=\sum_{k=0}^{\infty}\frac{k}{k!}=e$$
and
$$b_{13}=\sum_{k=0}^{\infty}\frac{k(k-1)}{2k!}=\frac{e}{2}$$
finally, the sum you want is:
$$\frac{1}{e}\left(3e+2e+\frac{e}{2}\right)=\frac{11}{2}$$
