Given,
2x=3y
⇒3x=2y
Now, let 3x=2y=k
\Rightarrow x=3k&y=2k
Now, given set A=1,2,3,.......,100
So, for k≤33 we get,
⇒R=(3,2),(6,4),(9,6),(12,8),....(99,66)
⇒n(R)=33
Now, R1 is symmetric relation on A and R⊂R1 so, we need elements (3,2),(6,4),(9,6),(12,8),....(99,66),(2,3),(4,6),(6,9),(8,12),....(66,99) in R1
Hence, the minimum number of elements in R1 is 33+33=66