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(D) An equivalence relation $R$ on a set $A$ is reflexive,symmetric,and transitive.$1$. Reflexive: For all $a \in A$,$(a, a) \in R$. Since $(a, a) \in

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(D) An equivalence relation $R$ on a set $A$ is reflexive,symmetric,and transitive.$1$. Reflexive: For all $a \in A$,$(a, a) \in R$. Since $(a, a) \in R$,then $(a, a) \in R^{-1}$,so $R^{-1}$ is reflexive.$2$. Symmetric: If $(a, b) \in R$,then $(b, a) \in R$. For $R^{-1}$,if $(a, b) \in R^{-1}$,then $(b, a) \in R$. Since $R$ is symmetric,$(a, b) \in R$,so $(b, a) \in R^{-1}$. Thus,$R^{-1}$ is symmetric.$3$. Transitive: If $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. For $R^{-1}$,if $(a, b) \in R^{-1}$ and $(b, c) \in R^{-1}$,then $(b, a) \in R$ and $(c, b) \in R$. Since $R$ is transitive,$(c, a) \in R$,which implies $(a, c) \in R^{-1}$. Thus,$R^{-1}$ is transitive.Since $R^{-1}$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,it is not the case that $R^{-1}$ is 'not' any of these.

If $R$ is an equivalence relation on a set $A$,then $R^{-1}$ is not :-

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