Let A be a symmetric matrix such that $\left|\mathrm{A}\right|=2\text{ and }\left[\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{3}{2}\end{array}\right]\mathrm{A}=\left[\begin{array}{cc}1& 2\\ \alpha & \beta \end{array}\right]$
If the sum of the diagonal elements of A is s, then $\frac{\mathrm{\beta s}}{{\mathrm{\alpha }}^2}$ is equal to _________.

$$\begin{gathered} \left|A\right|=2\left[\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{3}{2}\end{array}\right]A=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ \alpha \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\beta \end{array}\right] \\ Det on both sides \Rightarrow \beta -2\alpha =0 \end{gathered}$$
$Let A=\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c\end{array}\right]\Rightarrow \left|A\right|=ac-b^2=2\dots \dots \dots ..\left(i\right)$
$\begin{array}{l}\left[\begin{array}{ll}2& 1\\ 3& \frac{3}{2}\end{array}\right]\left[\begin{array}{ll}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]=\left[\begin{array}{cc}1& 2\\ \mathrm{\alpha }& 2\mathrm{\alpha }\end{array}\right]\\ 2\mathrm{a}+\mathrm{b}=1\dots \dots \left(\mathrm{ii}\right)\text{ and }2\mathrm{b}+\mathrm{c}=2\dots \dots \dots \left(\text{ iiii }\right)\\ 3\mathrm{a}+\frac{3\mathrm{b}}{2}=\mathrm{\alpha }\Rightarrow 3\mathrm{b}+\frac{3\mathrm{c}}{2}=2\mathrm{\alpha }\Rightarrow 6\mathrm{a}+3\mathrm{b}=3\mathrm{b}+\frac{3\mathrm{c}}{2}\\ \mathrm{c}=4\mathrm{a}\dots \dots .\left(\mathrm{iv}\right)\text{ from }\left(\mathrm{i}\right)\mathrm{\&}\left(\mathrm{ii}\right)\mathrm{a}=\frac{3}{4}\end{array}$
$\begin{array}{l}\mathrm{b}=-\frac{1}{2}\cdot \mathrm{c}=3\mathrm{S}=\mathrm{a}+\mathrm{c}=\frac{15}{4}\\ \mathrm{\alpha }=\frac{9}{4}+\left(-\frac{3}{4}\right)=\frac{6}{4}=\frac{3}{2}\Rightarrow \mathrm{\beta }=3\therefore \frac{\mathrm{\beta S}}{{\mathrm{\alpha }}^2}=5\end{array}$