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Let $A=\{2,3,6,8,9,11\}$ and $B=\{1,4,5,10,15\}$
Let $\mathrm{R}$ be a relation on $\mathrm{A} \times \mathrm{B}$ define by $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ if and only if $3 \mathrm{ad}-7 \mathrm{bc}$ is an even integer. Then the relation $\mathrm{R}$ is
reflexive but not symmetric.
transitive but not symmetric.
reflexive and symmetric but not transitive.
an equivalence relation.
Solution
$ \mathrm{A}=\{2,3,6,8,9,11\} \quad(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) $
$ \mathrm{B}=\{1,4,5,10,15\} \quad 3 \mathrm{ad}-7 \mathrm{bc} $
$ \text { Reflexive : }(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b})$
$\Rightarrow 3 \mathrm{ab}-7 \mathrm{ba}=-4 \mathrm{ab}$ always even so it is reflexive.
Symmetric : If $3 \mathrm{ad}-7 \mathrm{bc}=$ Even
Case $-I$ : odd odd
Case $-II$ : even even
$(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow 3 \mathrm{bc}-3 \mathrm{ab}$
Case $-I$ : odd odd
Case $-II$ : even even
so symmetric relation
Transitive :
Set $(3,4) R(6,4)$ Satisfy relation
Set $(6,4) R(3,1)$ Satisfy relation
but $(3,4) R(3,1)$ does not satisfy relation so not transitive.