Let A = \ 1, 2, 3, . ...100\ . Let R be a relation on A defined by...

Given,

$2x = 3y$

$⇒ \dfrac{x}{3} = \dfrac{y}{2}$

Now, let $\dfrac{x}{3} = \dfrac{y}{2} = k$

$⇒ x = 3k \text{ and } y = 2k$

Now, given set $A = \left\{1, 2, 3, . . . . . . . , 100\right\}$

So, for $k \le 33$ we get,

$⇒ R = \left\{\left(3, 2\right), \left(6, 4\right), \left(9, 6\right), \left(12, 8\right), . ... \left(99, 66\right)\right\}$

$⇒ n\left(R\right) = 33$

Now, ${R}_{1}$ is symmetric relation on $A$ and $R ⊂ {R}_{1}$ so, we need elements \left\{\left(3, 2\right), \left(6, 4\right), \left(9, 6\right), \left(12, 8\right), . ... \left(99, 66\right), \left(2, 3\right), \left(4, 6\right), \left(6, 9\right), \left(8, 12\right), . ... \left(66, 99\right)\right\} in ${R}_{1}$

Hence, the minimum number of elements in ${R}_{1}$ is $33 + 33 = 66$