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Question: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then \(R ^ { - 1 } o R\...

Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6);

(7, 6); (3, 7)} then R−1oRR ^ { - 1 } o R is

A

{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}

B

{(1, 1), (4, 4), (7, 7), (3, 3)}

C

{(1, 5), (1, 6), (3, 6)}

D

None of these

Answer

{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}

Explanation

Solution

We first find R−1R ^ { - 1 } we have

R−1={(5,4);(4,1);(6,4);(6,7);(7,3)}R ^ { - 1 } = \{ ( 5,4 ) ; ( 4,1 ) ; ( 6,4 ) ; ( 6,7 ) ; ( 7,3 ) \} we now obtain the

elements of R−1oRR ^ { - 1 } o R we first pick the element of R and then of R−1R ^ { - 1 } . Since (4,5)∈R( 4,5 ) \in R and (5,4)∈R−1( 5,4 ) \in R ^ { - 1 } , we have (4,4)∈R−1oR( 4,4 ) \in R ^ { - 1 } o R

Similarly,

(4,6)∈R,(6,4)∈R−1⇒(4,4)∈R−1oR( 4,6 ) \in R , ( 6,4 ) \in R ^ { - 1 } \Rightarrow ( 4,4 ) \in R ^ { - 1 } o R (4,6)∈R,(6,7)∈R−1⇒(4,7)∈R−1oR( 4,6 ) \in R , ( 6,7 ) \in R ^ { - 1 } \Rightarrow ( 4,7 ) \in R ^ { - 1 } o R (7,6)∈R,(6,4)∈R−1⇒(7,4)∈R−1oR( 7,6 ) \in R , ( 6,4 ) \in R ^ { - 1 } \Rightarrow ( 7,4 ) \in R ^ { - 1 } o R (7,6)∈R,(6,7)∈R−1⇒(7,7)∈R−1oR( 7,6 ) \in R , ( 6,7 ) \in R ^ { - 1 } \Rightarrow ( 7,7 ) \in R ^ { - 1 } o R

Hence R−1oR=R ^ { - 1 } o R ={(1, 1); (4, 4); (4, 7); (7 , 4), (7, 7); (3, 3)}..