Let $A$ be a symmetric matrix such that $|A|=2$ and $\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix}…

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 _________.

Solution

Let $A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$. And, $\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} A = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$. $\begin{aligned} \begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix} \end{aligned}$ $\begin{aligned} \begin{bmatrix} 2a+b & 2b+c \\ 3a+\frac{3b}{2} & 3b+\frac{3c}{2} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix} \end{aligned}$ So,

2a+b=12b+c=2

So, 4a-c=0   ....i

Also,

A=2ac-b2=2

a4a-1-2a2=2

4a2-4a2-1+4a=2

a=34

b=1-2a=1-64=-12

c=4a=3

And, 

3a+3b2=αα=94-34=323b+3c2=ββ=-32+92=3

$A=\begin{bmatrix} \frac{3}{4} & -\frac{1}{2} \\ -\frac{1}{2} & 3 \end{bmatrix}$ $Tr(A)=s=\frac{3}{4}+3=\frac{15}{4}$ $\frac{\beta \times s}{\alpha^{2}}=\frac{3 \times \frac{15}{4}}{\frac{9}{4}}=5$

Asked in: JEE Main 2023 (29 Jan Shift 2)

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