Let $ A= \left[ {\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} } \right] $ and $$e^A= I +A + \dfrac{A^2}{2!}+\dfrac{A^3}{3!}+...$$ If $e^A=[b_{ij}]$ then what is $$\dfrac{1}{e} \sum_{i=1}^{3} \sum_{j=1}^{3} b_{ij}$$
Can anyone suggest something how I should start? I can't think of anything.
Hint: Since $A=\operatorname{Id}+B$, where$$B=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$and since $\operatorname{Id}$ and $B$ commute, $e^A=e^{\operatorname{Id}}e^B$.