Consider the following relation R on the set | vedclass

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(A) For reflexive property:$(A, A) \in R$ because $A = I^{-1}AI$,where $I$ is the identity matrix,which is invertible. Thus,$R$ is reflexive.For symme

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Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is not a correct explanation for Statement-$1$.

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(A) For reflexive property:$(A, A) \in R$ because $A = I^{-1}AI$,where $I$ is the identity matrix,which is invertible. Thus,$R$ is reflexive.For symmetric property:If $(A, B) \in R$,then $A = P^{-1}BP$ for some invertible matrix $P$.Multiplying by $P$ on the left and $P^{-1}$ on the right: $PAP^{-1} = B$.Let $Q = P^{-1}$. Since $P$ is invertible,$Q$ is also invertible.Then $B = Q^{-1}AQ$,so $(B, A) \in R$. Thus,$R$ is symmetric.For transitive property:If $(A, B) \in R$ and $(B, C) \in R$,then $A = P^{-1}BP$ and $B = Q^{-1}CQ$ for some invertible matrices $P$ and $Q$.Substituting $B$: $A = P^{-1}(Q^{-1}CQ)P = (QP)^{-1}C(QP)$.Since $QP$ is invertible,$(A, C) \in R$. Thus,$R$ is transitive.Therefore,$R$ is an equivalence relation. Statement-$1$ is true.Statement-$2$ is a standard property of invertible matrices: $(MN)^{-1} = N^{-1}M^{-1}$. This is true.However,Statement-$2$ is a general property of matrix inversion and is not the specific reason why the relation $R$ (similarity) is an equivalence relation. Thus,Statement-$2$ is not the correct explanation for Statement-$1$.

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