Let R1, R2, R3, R4, be the relations defined on the set A = {1, 2, 3}. then
Match List-I with List-II
| List-I | List-II | ||
| Relation | Type | ||
| (A) | R1 = {(1, 1), (2, 2), (3, 3)} | (I) | Only transitive |
| (B) | R2 = {(1, 2), (2, 3), (1, 3)} | (II) | Equivalence relation |
| (C) | R3 = {(1, 2), (2,1), (1, 1), (2, 2)} | (III) | Reflexive and symmetric but not transitive |
| (D) | R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2), (1, 2), (2, 1)} | (IV) | Symmetric and transitive but not reflexive |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
On A = {1, 2, 3}, a reflexive relation must contain (1, 1), (2, 2) and (3, 3).
(A) R1 = {(1, 1), (2, 2), (3, 3)}
This matches (II) Equivalence relation.
It contains all three pairs (a, a), so it is reflexive. It has no pair (a, b) with a ≠ b, so the conditions for symmetry and transitivity can never fail.
(B) R2 = {(1, 2), (2, 3), (1, 3)}
This matches (I) Only transitive.
Not reflexive: (1, 1) is missing. Not symmetric: (1, 2) is in R2 but (2, 1) is not.
Transitive: the only chain is (1, 2), (2, 3), and (1, 3) is in R2.
(C) R3 = {(1, 2), (2,1), (1, 1), (2, 2)}
This matches (IV) Symmetric and transitive but not reflexive.
Not reflexive: (3, 3) is missing. Symmetric: (1, 2) and (2, 1) are both present.
Transitive: (1, 2), (2, 1) gives (1, 1); (2, 1), (1, 2) gives (2, 2); every other chain involves (1, 1) or (2, 2) and gives back a pair already present.
(D) R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2), (1, 2), (2, 1)}
This matches (III) Reflexive and symmetric but not transitive.
Reflexive: all of (1, 1), (2, 2), (3, 3) are present. Symmetric: (2, 3), (3, 2) and (1, 2), (2, 1) come in pairs.
Not transitive: (1, 2) and (2, 3) are present but (1, 3) is not.
So the matching is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).