Suppose A = {1, 2, 3}.
For S1 :
Suppose R = {(1,1), (2,2), (3,3), (1,2)}.
This relation R is reflexive.
Now, $R^{-1}$ = {(1,1), (2,2), (3,3), (2,1)}.
$R^{-1}$ $\neq$ R even when R is reflexive.
Thus, S1 is False.
For S2 :
Assume, R is symmetric ie $\forall$a,b aRb $\implies$ bRa.
aRb $\implies$ b$R^{-1}$a and bRa $\implies$ a$R^{-1}$b.
Therefore, (in simpler terms) all elements of R will form pair like (a,b) and (b,a) due to symmetry and all elements of $R^{-1}$ will have pairs like (b,a) and (a,b) for ever pair in R.
This makes R = $R^{-1}$.
Now, assume R = $R^{-1}$.
Suppose, (a,b) $\in$ R $\implies$ (b,a) $\in$ $R^{-1}$.
But R = $R^{-1}$ is given $\implies$ (b,a) $\in$ R and (a,b) $\in$ $R^{-1}$.
Now, we have (a,b) $\in$ R $\implies$ (b,a) $\in$ R.
This makes R symmetric.
Thus, S2 is True.
Answer :- B.
