JEE Main · Mathematics · Sets and Relations
C. { ( 3, 3 ) , ( 3, 5 ) , ( 5, 3 ) , ( 5, 5 )
A=1,2,3,4A = { 1, 2, 3, 4 }A=1,2,3,4~and~B=1,3,5B = { 1, 3, 5 }B=1,3,5∴R=(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)∴ R = { ( 1, 3 ) , ( 1, 5 ) , ( 2, 3 ) , ( 2, 5 ) , ( 3, 5 ) , ( 4, 5 ) }∴R=(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)and~R−1={(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}R^{- 1} = \left\{( 3, 1 ) , ( 5, 1 ) , ( 3, 2 ) , ( 5, 2 ) , ( 5, 3 ) , ( 5, 4 )\right\}R−1={(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}Now,~3→R−11→R3⇒(3,3)∈ROR−13\overset{R- 1}{→}1\overset{R}{→}3 ⇒ \left(3, 3\right) \in ROR^{- 1}3→R−11→R3⇒(3,3)∈ROR−15→R−11→R5⇒(5,5)∈ROR−15\overset{R- 1}{→}1\overset{R}{→}5 ⇒ \left(5, 5\right) \in ROR^{- 1}5→R−11→R5⇒(5,5)∈ROR−15→R−12→R3⇒(5,3)∈ROR−15\overset{R- 1}{→}2\overset{R}{→}3 ⇒ \left(5, 3\right) \in ROR^{- 1}5→R−12→R3⇒(5,3)∈ROR−13→R−11→R5⇒(3,5)∈ROR−13\overset{R- 1}{→}1\overset{R}{→}5 ⇒ \left(3, 5\right) \in ROR^{- 1}3→R−11→R5⇒(3,5)∈ROR−1∴ROR−1={(3,3),(3,5),(5,3),(5,5)}∴ ROR^{- 1} = \left\{\left(3, 3\right), \left(3, 5\right), \left(5, 3\right), \left(5, 5\right)\right\}∴ROR−1={(3,3),(3,5),(5,3),(5,5)}
A=1,2,3,4A = { 1, 2, 3, 4 }A=1,2,3,4~and~B=1,3,5B = { 1, 3, 5 }B=1,3,5
∴R=(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)∴ R = { ( 1, 3 ) , ( 1, 5 ) , ( 2, 3 ) , ( 2, 5 ) , ( 3, 5 ) , ( 4, 5 ) }∴R=(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)
and~R−1={(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}R^{- 1} = \left\{( 3, 1 ) , ( 5, 1 ) , ( 3, 2 ) , ( 5, 2 ) , ( 5, 3 ) , ( 5, 4 )\right\}R−1={(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}
Now,~
3→R−11→R3⇒(3,3)∈ROR−13\overset{R- 1}{→}1\overset{R}{→}3 ⇒ \left(3, 3\right) \in ROR^{- 1}3→R−11→R3⇒(3,3)∈ROR−1
5→R−11→R5⇒(5,5)∈ROR−15\overset{R- 1}{→}1\overset{R}{→}5 ⇒ \left(5, 5\right) \in ROR^{- 1}5→R−11→R5⇒(5,5)∈ROR−1
5→R−12→R3⇒(5,3)∈ROR−15\overset{R- 1}{→}2\overset{R}{→}3 ⇒ \left(5, 3\right) \in ROR^{- 1}5→R−12→R3⇒(5,3)∈ROR−1
3→R−11→R5⇒(3,5)∈ROR−13\overset{R- 1}{→}1\overset{R}{→}5 ⇒ \left(3, 5\right) \in ROR^{- 1}3→R−11→R5⇒(3,5)∈ROR−1
∴ROR−1={(3,3),(3,5),(5,3),(5,5)}∴ ROR^{- 1} = \left\{\left(3, 3\right), \left(3, 5\right), \left(5, 3\right), \left(5, 5\right)\right\}∴ROR−1={(3,3),(3,5),(5,3),(5,5)}