Which relation on A={1,2,3} is symmetric but not transitive?

Correct answer: R={(1,2),(2,1),(1,3),(3,1)}

Correct answer: A. A relation is symmetric when every pair appears with its reverse. In option A, (1,2) is paired with (2,1), and (1,3) with (3,1), so symmetry holds. To test transitivity, use (2,1) and (1,3). Transitivity would require (2,3), but (2,3) is not in R. Hence option A is not transitive. Option B is the identity relation: whenever two pairs can be composed, the required pair is already present, so it is transitive. Option C is not symmetric because (1,2) is present but (2,1) is absent; therefore it cannot be the answer, regardless of its other property. Option D is the universal relation, containing every ordered pair, so it is both symmetric and transitive. The key strategy is to disprove transitivity with one chain xRy and yRz for which xRz is missing; one such counterexample is enough.