Let A be a symmetric matrix such that $\mathrm{|A|=2}$ and $$\left[ {\matrix{ 2 & 1 \cr 3 & {{3 \over 2}} \cr } } \right]A = \left[ {\matrix{ 1 & 2 \cr \alpha & \beta \cr } } \right]$$. If the sum of the diagonal elements of A is $s$, then $\frac{\beta s}{\alpha^2}$ is equal to __________.

<p>$A = \left( {\matrix{ a & c \cr c & b \cr } } \right)$</p> <p>$|A| = ab - {c^2} = 2$ ...... (1)</p> <p>$$\left( {\matrix{ 2 & 1 \cr 3 & {{3 \over 2}} \cr } } \right)\left( {\matrix{ a & c \cr c & b \cr } } \right) = \left( {\matrix{ 1 & 2 \cr \alpha & \beta \cr } } \right)$$</p> <p>$2a + c = 1$ ..... (2)</p> <p>$2c + b = 2$ ..... (3)</p> <p>$3a + {3 \over 2}c = \alpha$ .... (4)</p> <p>$3c + {3 \over 2}b = \beta$ ..... (5)</p> <p>From (1), (2) and (3)</p> <p>$a = {3 \over 4},b = 3,c = - {1 \over 2}$</p> <p>$\Rightarrow$ Now $\alpha = {6 \over 4}$</p> <p>$\beta = 3$</p> <p>$s = {{15} \over 4}$</p> <p>$${{\beta s} \over {{\alpha ^2}}} = {{3 \times {{15} \over 4}} \over {{{\left( {{6 \over 4}} \right)}^2}}} = {{{{45} \over 4}} \over {{9 \over 4}}} = 5$$</p>