2. Let the set
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{5}\right\}\left(a_{i} \in \mathbf{R}_{+}, i=1,2, \cdots, 5\right) .
$$

If the set of the products of the four elements in all four-element subsets of set $A$ is $B=\{2,3,4,6,9\}$, then the sum of the elements in set $A$ is $\qquad$
2. $\frac{49}{6}$.

Obviously, in all the four-element subsets of set $A$, each element appears 4 times.
Then $\left(a_{1} a_{2} \cdots a_{5}\right)^{4}=1296 \Rightarrow a_{1} a_{2} \cdots a_{5}=6$.
Thus, the products of the four elements of set $A$ are 2, 3, $\frac{2}{3}$, 1, $\frac{3}{2}$, and their sum is $\frac{49}{6}$.