JEE Main 2023 — Mathematics Algebra

Let

$$A=[\begin{matrix}a & b \\ b & c\end{matrix}]$$A=[ab​bc​]

And,

$$[\begin{matrix}2 & 1 \\ 3 & \frac{3}{2}\end{matrix}]A=[\begin{matrix}1 & 2 \\ \alpha & \beta \end{matrix}]$$[23​123​​]A=[1α​2β​]

$$\Rightarrow [\begin{matrix}2 & 1 \\ 3 & \frac{3}{2}\end{matrix}][\begin{matrix}a & b \\ b & c\end{matrix}]=[\begin{matrix}1 & 2 \\ \alpha & \beta \end{matrix}]$$⇒[23​123​​][ab​bc​]=[1α​2β​]

$$\Rightarrow [\begin{matrix}2a+b & 2b+c \\ 3a+\frac{3b}{2} & 3b+\frac{3c}{2}\end{matrix}]=[\begin{matrix}1 & 2 \\ \alpha & \beta \end{matrix}]$$⇒[2a+b3a+23b​​2b+c3b+23c​​]=[1α​2β​]

So,

$$2a+b=1 2b+c=2$$2a+b=12b+c=2

So, $4a-c=0....(i)$4a−c=0....(i)

Also,

$$|A|=2\Rightarrow ac-{b}^{2}=2$$∣A∣=2⇒ac−b2=2

$$\Rightarrow a(4a)-{(1-2a)}^{2}=2$$⇒a(4a)−(1−2a)2=2

$$\Rightarrow 4{a}^{2}-4{a}^{2}-1+4a=2$$⇒4a2−4a2−1+4a=2

$$\Rightarrow a=\frac{3}{4}$$⇒a=43​

$$\Rightarrow b=1-2a=1-\frac{6}{4}=-\frac{1}{2}$$⇒b=1−2a=1−46​=−21​

$$c=4a=3$$c=4a=3

And,

$$3a+\frac{3b}{2}=\alpha \Rightarrow \alpha =\frac{9}{4}-\frac{3}{4}=\frac{3}{2} 3b+\frac{3c}{2}=\beta \Rightarrow \beta =-\frac{3}{2}+\frac{9}{2}=3$$3a+23b​=α⇒α=49​−43​=23​3b+23c​=β⇒β=−23​+29​=3

So,

$$A=[\begin{matrix}\frac{3}{4} & -\frac{1}{2} \\ -\frac{1}{2} & 3\end{matrix}]$$A=[43​−21​​−21​3​]

So, $\mathrm{Tr}(A)=s=\frac{3}{4}+3=\frac{15}{4}$Tr(A)=s=43​+3=415​

$$\frac{\beta \times s}{{\alpha }^{2}}=\frac{3\times \frac{15}{4}}{\frac{9}{4}}=5$$α2β×s​=49​3×415​​=5