Let A be a symmetric matrix such that |A| = 2 and [ cases 2 & 1 3...

Let

A = \left[\begin{cases}a & b \\ b & c\end{cases}\right]

And,

\left[$\begin{cases}2 & 1 \\ 3 & \dfrac{3}{2}\end{cases}$\right]A = \left[$\begin{cases}1 & 2 \\ \alpha & \beta\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}2 & 1 \\ 3 & \dfrac{3}{2}\end{cases}$\right]\left[$\begin{cases}a & b \\ b & c\end{cases}$\right] = \left[$\begin{cases}1 & 2 \\ \alpha & \beta\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}2a + b & 2b + c \\ 3a + \dfrac{3b}{2} & 3b + \dfrac{3c}{2}\end{cases}$\right] = \left[$\begin{cases}1 & 2 \\ \alpha & \beta\end{cases}$\right]$</p><p>So,</p><p>$2a + b = 12b + c = 2$</p><p>So,$4a - c = 0 . . . . \left(i\right)$</p><p>Also,</p><p>\left|A\right| = 2 ⇒ ac - {b}^{2} = 2$

$⇒ a\left(4a\right) - {\left(1 - 2a\right)}^{2} = 2$

$⇒ 4{a}^{2} - 4{a}^{2} - 1 + 4a = 2$

$⇒ a = \dfrac{3}{4}$

$⇒ b = 1 - 2a = 1 - \dfrac{6}{4} = - \dfrac{1}{2}$

$c = 4a = 3$

And,

$3a + \dfrac{3b}{2} = \alpha ⇒ \alpha = \dfrac{9}{4} - \dfrac{3}{4} = \dfrac{3}{2}3b + \dfrac{3c}{2} = \beta ⇒ \beta = - \dfrac{3}{2} + \dfrac{9}{2} = 3$

So,

A = \left[\begin{cases}\dfrac{3}{4} & - \dfrac{1}{2} \\ - \dfrac{1}{2} & 3\end{cases}\right]

So, $Tr\left(A\right) = s = \dfrac{3}{4} + 3 = \dfrac{15}{4}$

$\dfrac{\beta \times s}{{\alpha}^{2}} = \dfrac{3 \times \dfrac{15}{4}}{\dfrac{9}{4}} = 5$