• # question_answer Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6);     (7, 6); (3, 7)} then ${{R}^{-1}}oR$ 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

We first find${{R}^{-1}},$ we have ${{R}^{-1}}=\{(5,\,4);\,(4,\,1);\,(6,\,4);\,(6,\,7);\,(7,\,3)\}$. We now obtain the elements of ${{R}^{-1}}oR$ we first pick the element of R and then of ${{R}^{-1}}$. Since $(4,\,5)\in R$ and $(5,\,4)\in {{R}^{-1}}$, we have $(4,\,4)\in {{R}^{-1}}oR$ Similarly, $(1,\,4)\in R,\,(4,\,1)\in {{R}^{-1}}\Rightarrow \,(1,\,1)\in {{R}^{-1}}oR$ $(4,\,6)\in R,\,(6,\,4)\in {{R}^{-1}}\Rightarrow \,(4,\,4)\in {{R}^{-1}}oR,$   $(4,\,6)\in R,\,(6,\,7)\in {{R}^{-1}}\Rightarrow \,(4,\,7)\in {{R}^{-1}}oR$ $(7,\,6)\in R,\,(6,\,4)\in {{R}^{-1}}\Rightarrow \,(7,\,4)\in {{R}^{-1}}oR,$        $(7,\,6)\in R,\,(6,\,7)\in {{R}^{-1}}\Rightarrow \,(7,\,7)\in {{R}^{-1}}oR$ $(3,\,7)\in R,\,(7,\,3)\in {{R}^{-1}}\Rightarrow \,(3,\,3)\in {{R}^{-1}}oR,$ Hence,${{R}^{-1}}oR=${(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.