QJEEM23Mac5ae: JEE Main 2023 Mathematics - Class 12 JEEM23Mac5ae Solution: Let A be a symmetric matrix such that and [ 2 1 3 3 2 ] A = [ 1 …

$$A=\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$$

$|A| = ab - {c^2} = 2$ ...... (1)

$$\left(\begin{array}{cc}2& 1\\ 3& \frac{3}{2}\end{array}\right)\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)=\left(\begin{array}{cc}1& 2\\ \alpha & \beta \end{array}\right)$$

$2a + c = 1$ ..... (2)

$2c + b = 2$ ..... (3)

$3a + {3 \over 2}c = \alpha$ .... (4)

$3c + {3 \over 2}b = \beta$ ..... (5)

From (1), (2) and (3)

$a = {3 \over 4},b = 3,c = - {1 \over 2}$

$\Rightarrow$ Now $\alpha = {6 \over 4}$

$\beta = 3$

$s = {{15} \over 4}$

${{\beta s} \over {{\alpha ^2}}} = {{3 \times {{15} \over 4}} \over {{{\left( {{6 \over 4}} \right)}^2}}} = {{{{45} \over 4}} \over {{9 \over 4}}} = 5$