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## Here's the Solution to this Question

1.The matrix can be written in relation form,

a) To be reflexive relation should satisfy

(a,a) for all elements i.e all diagonal elemtents must be 1

hence the relation is Not Reflexive.

b)It is Not Irreflexive .

To be irreflexive all diagonal elements must be 0

But (1,1) is present.

c) To be symmetric it should satisfy

if(a, b) is present then (b,a) must me present

It does not satisfy for (1,7) because (7,1) is not present hence the relation is not symmetric.

d) To be asymmetric

if(a, b) is present then (b, a) should not be present.

but (1,1) is present therefore Not Asymmetric.

e) To be antisymmetric

if(a,b) and (b,a) is present then a must be equal to b

hence the relation satisfy and it is Antisymmetric.

f)To be in transitive it should satisfy

if (a,b ) and ( b,c) is present then (a,c) must be present.

But here for no element this property is satisfied hence it is Not Transitive.

2.For A={1,2,3,4,5}

The Relation R is given as-

R={(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,%),(3,2),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5)}

(a) Diagraph is- (b)These are paths of length 2 from vertex 3.

$P_1:3\to 2 \to 5 \\ P_2:3\to 2 \to 4 \\ P_3:3\to 2 \to 3\\ P_4:3\to 4 \to 5\\ \\$

(c) Domain (R)={1,2,3,4,5}

Range (R)={2,3,4,5}

By definition

Domain (R)={$x\in A:(x,y)\in R$ }

Range (R)={$y\in A:(x,y)\in R$ }

(d) R={$(a,b)\in A\times A:\dfrac{a}{b} }

{(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,%),(3,2),(3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5)}