Let a relation R be defined by R = {(4, 5), ( | vedclass

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(A) First,we find ${R^{ - 1}}$ by reversing the ordered pairs in $R$: ${R^{ - 1}} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\}$. Now,we find the compo

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

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(A) First,we find ${R^{ - 1}}$ by reversing the ordered pairs in $R$: ${R^{ - 1}} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\}$. Now,we find the composition ${R^{ - 1}}oR$ by checking all pairs $(x, y) \in R$ and $(y, z) \in {R^{ - 1}}$ such that $(x, z) \in {R^{ - 1}}oR$: $1$. $(4, 5) \in R$ and $(5, 4) \in {R^{ - 1}} \Rightarrow (4, 4) \in {R^{ - 1}}oR$. $2$. $(1, 4) \in R$ and $(4, 1) \in {R^{ - 1}} \Rightarrow (1, 1) \in {R^{ - 1}}oR$. $3$. $(4, 6) \in R$ and $(6, 4) \in {R^{ - 1}} \Rightarrow (4, 4) \in {R^{ - 1}}oR$. $4$. $(4, 6) \in R$ and $(6, 7) \in {R^{ - 1}} \Rightarrow (4, 7) \in {R^{ - 1}}oR$. $5$. $(7, 6) \in R$ and $(6, 4) \in {R^{ - 1}} \Rightarrow (7, 4) \in {R^{ - 1}}oR$. $6$. $(7, 6) \in R$ and $(6, 7) \in {R^{ - 1}} \Rightarrow (7, 7) \in {R^{ - 1}}oR$. $7$. $(3, 7) \in R$ and $(7, 3) \in {R^{ - 1}} \Rightarrow (3, 3) \in {R^{ - 1}}oR$. Combining these,we get ${R^{ - 1}}oR = \{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)\}$.

Let a relation $R$ be defined by $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then ${R^{ - 1}}oR$ is

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