laplace form and matrix exponential Given the matrix $(I-A)^{-1}$ and $B$, can we compute $e^{A+B}$, where $e^X$ is defined to be $\sum_{i=0}^{\infty} \frac{X^i}{i!}$.
(Note that $A$ and $B$ do not commute, and hence $e^A \cdot e^B \neq e^{A+B}$).
Now I've observed that Laplace transformation might be a useful tool. I've obtained that
$$\mathcal{L}[e^{tA+B}](s) ={(sI-A)}^{-1}e^{B}.$$
So is the above (inverse) laplace transformation really useful to compute $e^{A+B}$ from  $(I-A)^{-1}$ and $B$? How can I get the resultant $e^{A+B}$ from the Laplace transformation?
Hope anyone who is familiar with linear algebra and Laplace transformation could give me a hand. Thanks!

$e^{A+B}$ is not uniquely determined by $e^A$ and $e^B$.
First take $A = \left[ \begin{array}{cc} 0 & -\pi \\\ \pi & 0 \end{array} \right]$ and $B = \left[ \begin{array}{cc} \pi & 0 \\\ 0 & -\pi \end{array} \right]$. Then $A + B$ squares to zero, so we have
$$e^A = \left[ \begin{array}{cc} \cos \pi & - \sin \pi \\\ \sin \pi & \cos \pi \end{array} \right] = \left[ \begin{array}{cc} -1 & 0 \\\ 0 & -1 \end{array} \right], e^B = \left[ \begin{array}{cc} e^{\pi} & 0 \\\ 0 & e^{-\pi} \end{array} \right], e^{A+B} = \left[ \begin{array}{cc} 1 + \pi & -\pi \\\ \pi & 1 - \pi \end{array} \right].$$
Now replace $A$ with $\left[ \begin{array}{cc} 0 & - 3\pi \\\ 3\pi & 0 \end{array} \right]$. Then $e^A$ is the same, but now
$$A + B = \left[ \begin{array}{cc} \pi & - 3\pi \\\ 3 \pi & - \pi \end{array} \right]$$
has eigenvalues $\pm \pi i \sqrt{7}$, so the eigenvalues of $e^{A+B}$ are different from what they were before. 
