A = {1,2, 3}
R = {(1.2), (2, 1)}
Since (a, a) ∈ R ∀ a ∈ A R is not reflexive Now (1, 2) ∈ R ⇒ (2, 1) ∈ R and (2, 1) ∈ R ⇒ (1,2) ∈ R ∴ (a, b) ∈ R ⇒ (b, a) ∈ R ∀ (a, b) ∈ R ∴ R is symmetric Again (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R ∴ R is not transitive.

(i) Let A = {1, 2}.
Then A x A = {(1,1), (1,2), (2,1), (2,2) }.
Let R = {(1,2), (2,1 )} .
Then R ⊆ A x A and hence R is a relation on the set A.
R is symmetric since (a, b) ∈ R ⇒ (b. a) ∈ R.
R is not reflexive since I ∈ A but (1,1) ∉ R.
R is not transitive since (1, 2) ∈ R, (2,1) ∈ R but (1,1) ∉ R.
(ii) Let A = {1,2,3}
Then A x A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.
Let R = {(1,1), (2,2), (1,2), (2,1), (1,3), (2,3)}.
Then R is transitive since (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R.
R is not reflexive since 3 G A but (3.3) ∉ R.
R is not symmetric since (1,3) ∈R but (3,1) ∉ R.
(iii) Let A = {1,2 3}
Then A x A = {(1, 1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) }.
Let R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}.
R is a relation on A as R ⊆ A x A.
R is reflexive as (a, a) ∈ R ∀ a ∈ A.
Also. R is symmetric since (a. b) ∈ R implies that (b, a) ∈R.
But R is not transitive since (1,2) ∈R arid (2,3) ∈R but (1,3) ∢ R.
(iv) Let A = {1,2,3}.
Then A x A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.
Let R = {(1,1), (2,2), (3,3), (1,3)}.
Then R is a relation on A as R ⊆ A x A.
R is reflexive since (a, a) ∈R ∀ a ∈ A.
R is not symmetric as (1,3) ∈R and (3,1) ∉ R. R is transitive since (a, b) ∈R and (b, c) ∈R implies that (a, c) ∈R.
(v) Let A = {1,2,3}
Then A x A = {(1,1), (1,2), (1, 3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
Let R = {(1,1), (1,2), (2,1), (2,2)}.
R is not reflexive as 3∈ A and (3,3) ∉ R.
R is symmetric as (a, b) ∈ R ⇒ (b, a) ∈R.
R is transitive since (a, b) ∈ R and (b, c) ∈R implies that (a, c) ∈ R.

Since the school is boys school i.e. there is not girl student no student of the school can be sister of any student of the school.
∴ R ⇒ R is the empty relation.
Clearly the difference between heights of a and b is less than 3 metres.
∴ R’ = A x A is the universal relation.

R= {(L₁, L₂) : L₁ is perpendicular to L₂} Since no line can be perpendicular to itself ∴ R is not reflexive.
Let (L₁, L₂) ∈ R
∴ L₁ is perpendicular to L₂ ⇒ L₂ is peipendicular to L₁
⇒ (L₂, L₁) ∈ R
∴ (L₁L₂) ∈ R ⇒ (L₂, L₁) ∈ R
∴ R is symmetric
Again we know that if L₁ is perpendicular to L₂ and L₂ is perpendicular to L₃, then L₁can never be perpendicular to L₃.
∴ (L₁, L₂) ∈ R, (L₂, L₃) ∈ R does not imply (L₁, L₃) ∈ R
∴ R is not transitive.

Let A = {1. 2, 3}.
Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
Here R is reflexive as (a, a)∈ R ∀ a ∈ A.
R is not symmetric as (1, 2) ∈ R but (2, 1) ∈ R.
R is not transitive as (1.2) ∈ R and (2, 3) ∈ R but(1,3) ∈ R.
∴ R is a relation which is reflexive but neither symmetric not transitive.