Let A be a symmetric matrix such that $$|A| = 2$$ and $$\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} A = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$$. If the sum of the diagonal elements of A is $$s$$, then $$\frac{\beta s}{\alpha^2}$$ is equal to ______.

Let $$A = \begin{bmatrix} p & q \\ q & r \end{bmatrix}$$ (since $$A$$ is symmetric) with $$|A| = pr - q^2 = 2$$.

We have $$\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} \begin{bmatrix} p & q \\ q & r \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$$.

Multiplying the matrices yields $$\begin{bmatrix} 2p + q & 2q + r \\ 3p + \frac{3q}{2} & 3q + \frac{3r}{2} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$$.

From the first row we have $$2p + q = 1$$ and $$2q + r = 2$$, so $$q = 1 - 2p$$ and $$r = 2 - 2q = 2 - 2(1 - 2p) = 4p$$.

Using the condition $$|A| = pr - q^2 = 2$$ gives $$p(4p) - (1 - 2p)^2 = 2$$, which simplifies to $$4p^2 - 1 + 4p - 4p^2 = 2$$ and hence $$4p - 1 = 2$$, so $$p = \frac{3}{4}$$. Therefore $$q = -\frac{1}{2}$$ and $$r = 3$$.

From the second row we find $$\alpha = 3 \cdot \frac{3}{4} + \frac{3}{2} \cdot \left(-\frac{1}{2}\right) = \frac{9}{4} - \frac{3}{4} = \frac{3}{2}$$ and $$\beta = 3 \cdot \left(-\frac{1}{2}\right) + \frac{3}{2} \cdot 3 = -\frac{3}{2} + \frac{9}{2} = 3$$.

The sum of diagonal elements is $$s = \frac{3}{4} + 3 = \frac{15}{4}$$, and thus $$\frac{\beta s}{\alpha^2} = \frac{3 \cdot \frac{15}{4}}{\left(\frac{3}{2}\right)^2} = \frac{\frac{45}{4}}{\frac{9}{4}} = \frac{45}{9} = \boxed{5}$$.