we can also check like this-
@Prateek Raghuvanshi $R=(x,y)$ ∈ $N^{2}$ here means $R=(x,y)$ ∈ $N\times N$ not $R=(x,y)$ ∈ $N^{2} \times N^{2}$ as suggested by your solution.
Anti
@Ayush You are correct about reflexivity and symmetry. The main diagonal is all 1’s, so the relation is reflexive, and the matrix is symmetric about the main diagonal, so the relation is symmetric.
As for
@sripo (4,16) and (16,4) both belong to R, since these are symmetric pairs and 4 !=16 which violates antisymmetric property therefore R is not antisymmetric.
And For Transitive : Both (4,16) and (16,4) belongs to R therefore (4,4) must also belong to R but since here
"x !=y" is the condition therefore (4,4) is not allowed and it is not Transitive as well.
@sripo
Relation should be {(1,4)(4,1)...}. (According to the given question)
$\because (x,y) \in N^2$