Mathematics - Matrices Question with Solution | TestHub

Let

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

And,

$\left[\begin{array}{cc}2& 1\\ 3& \frac{3}{2}\end{array}\right]A=\left[\begin{array}{cc}1& 2\\ \alpha & \beta \end{array}\right]$

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

$\Rightarrow \left[\begin{array}{cc}2a+b& 2b+c\\ 3a+\frac{3b}{2}& 3b+\frac{3c}{2}\end{array}\right]=\left[\begin{array}{cc}1& 2\\ \alpha & \beta \end{array}\right]$

So,

$$\begin{gathered} 2a+b=1 \\ 2b+c=2 \end{gathered}$$

So, $4a-c=0 ....\left(\mathrm{i}\right)$

Also,

$\left|A\right|=2\Rightarrow ac-b^2=2$

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

$\Rightarrow 4a^2-4a^2-1+4a=2$

$\Rightarrow a=\frac{3}{4}$

$\Rightarrow b=1-2a=1-\frac{6}{4}=-\frac{1}{2}$

$c=4a=3$

And, 

$$\begin{gathered} 3a+\frac{3b}{2}=\alpha \Rightarrow \alpha =\frac{9}{4}-\frac{3}{4}=\frac{3}{2} \\ 3b+\frac{3c}{2}=\beta \Rightarrow \beta =-\frac{3}{2}+\frac{9}{2}=3 \end{gathered}$$

So,

$A=\left[\begin{array}{cc}\frac{3}{4}& -\frac{1}{2}\\ -\frac{1}{2}& 3\end{array}\right]$

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

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