Let A = [ a _ ij ] _ 2 × 2 , where a ≠ ij 0 for all i, j and A ^ 2...

Let,

A = \left[\begin{cases}p & q \\ r & s\end{cases}\right]

{A}^{2} = \left[\begin{cases}p & q \\ r & s\end{cases}\right]\left[\begin{cases}p & q \\ r & s\end{cases}\right] = \left[\begin{cases}{p}^{2} + qr & pq + qs \\ rp + rs & qr + {s}^{2}\end{cases}\right]

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

⇒ \left[\begin{cases}{p}^{2} + qr & pq + qs \\ rp + rs & qr + {s}^{2}\end{cases}\right] = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right]

Now on comparing both side we get,

$⇒ {p}^{2} + qr = 1, q ( p + s ) = 0$

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

Now on solving above relation we get,

$q \ne 0 ⇒ p + s = 0 ⇒ a = 0$

And $b = | A | = ps - qr = - {p}^{2} - qr = - 1 ( ∵ s = - p )$

$∴ 3{a}^{2} + 4{b}^{2} = 4$