Trace of the exponential of a diagonal matrix

Let $A=\begin{bmatrix}3&0\\0&6\end{bmatrix}$. The trace of the matrix exponential $e^A$ can be written as $e^a+e^b$ with integers $a,b$. What is $ab$?

$18$. For a diagonal matrix the power series $\sum A^k/k!$ acts entry by entry on the diagonal, so $e^A=\operatorname{diag}(e^3,e^6)$ and its trace is $e^3+e^6$; hence $ab=3\cdot6=18$.