Given $A = A ^{-1}$

$\therefore A ^{2}= A \cdot A ^{-1}= I$

${\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] }$

$\Rightarrow\left[\begin{array}{ll} a ^{2}+ bc & ab + bd \\ ac + cd & bc + d ^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\therefore a ^{2}+ bc =1$

$ab + bd =0$

$ac + cd =0$

$bc + d ^{2}=1$

$(1) - (4) gives$

$a^{2}-d^{2}=0$

$(a+d)=0 \text { or } a-d=0$

Case$-I$

$a+d=0 \Rightarrow(a, d)=(-1,1),(0,0),(1,-1)$

(a) $(a, d)=(-1,1)$

$\therefore$ from equation (1)

$1+ bc =1 \Rightarrow bc =0$

$b =0 C =12$ possibilities

$c =0 b =12$ possibilities

but $(0,0)$ is repeated

$\therefore 2 \times 12=24$

$24-1($ repeated $)=23$ pairs

$(b)$ $(a, d)=(1,-1) \Rightarrow b c=0 \rightarrow 23$ pairs

$(c)$ $( a , d )=(0,0) \Rightarrow bc =1$

$(b, c)=(1,1) \&(-1,-1), 2$ pairs

Case $- II$

$a = d$

from $(2)$ and $(3)$

$a \neq 0$ then $b=c=0$

$a ^{2}=1$

$a =\pm 1= d$

$( a , d )=(1,1),(-1,-1) \rightarrow 2$ pairs

$\therefore$ Total $=23+23+2+2$

$=50$ pairs