Let A = a i j 2 × 2 , where a ≠ i j 0 for all i , j and A 2 = I , Let a be the sum of all…

Let,

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

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

Now given, $A^2=I$

$\Rightarrow \left[\begin{array}{ll}{p}^2+qr& pq+qs\\ rp+rs& qr+s^2\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$

Now on comparing both side we get,

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

And $r(p+s)=0, qr+s^2=1$

Now on solving above relation we get,

$q\ne 0\Rightarrow p+s=0\Rightarrow a=0$

And $b=\left|A\right|=ps-qr=-p^2-qr=-1(\because s=-p)$

$\therefore 3a^2+4b^2=4$