Question

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

Answer 1

Answer: 5

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