JEE Main 2023 — Mathematics Algebra

Let,

$$A=[\begin{matrix}p & q \\ r & s\end{matrix}]$$A=[pr​qs​]

$${A}^{2}=[\begin{matrix}p & q \\ r & s\end{matrix}][\begin{matrix}p & q \\ r & s\end{matrix}]=[\begin{matrix}{p}^{2}+qr & pq+qs \\ rp+rs & qr+{s}^{2}\end{matrix}]$$A2=[pr​qs​][pr​qs​]=[p2+qrrp+rs​pq+qsqr+s2​]

Now given, ${A}^{2}=I$A2=I

$$\Rightarrow [\begin{matrix}{p}^{2}+qr & pq+qs \\ rp+rs & qr+{s}^{2}\end{matrix}]=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$⇒[p2+qrrp+rs​pq+qsqr+s2​]=[10​01​]

Now on comparing both side we get,

$$\Rightarrow {p}^{2}+qr=1,q(p+s)=0$$⇒p2+qr=1,q(p+s)=0

And $r(p+s)=0,qr+{s}^{2}=1$r(p+s)=0,qr+s2=1

Now on solving above relation we get,

$$q\neq 0\Rightarrow p+s=0\Rightarrow a=0$$q=0⇒p+s=0⇒a=0

And $b=|A|=ps-qr=-{p}^{2}-qr=-1(\because s=-p)$b=∣A∣=ps−qr=−p2−qr=−1(∵s=−p)

$$\therefore 3{a}^{2}+4{b}^{2}=4$$∴3a2+4b2=4