Question

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

Answer 1

$\begin{bmatrix}2 & 1 \\ 3 & \frac{3}{2}\end{bmatrix}\begin{bmatrix}a & b \\ b & c\end{bmatrix}=\begin{bmatrix}1 & 2 \\ \alpha & \beta\end{bmatrix}$
Now $a c-b^2=2$ and $2 a+b=1$
and $2 b+c=2$
solving all these above equations we get
$\frac{1-b}{2} \times\left(\frac{2-2 b}{1}\right)-b^2=2 $
$ \Rightarrow(1-b)^2-b^2=2 $
$ \Rightarrow 1-2 b=2$
$ \Rightarrow b=-\frac{1}{2} \text { and } a=\frac{3}{4} \text { and } c=3$
Hence $\alpha=3 a +\frac{3 b }{2}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}$
and $\beta=3 b+\frac{3 c}{2}=-\frac{3}{2}+\frac{9}{2}=3$
also $s = a + c =\frac{15}{4}$
$\therefore \frac{\beta s}{\alpha^2}=\frac{3 \times 15}{4 \times \frac{9}{4}}=5$