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(D) Let the relation be defined as $R = \{(A, B) : B = P A Q^{-1}\}$ where $P$ and $Q$ are invertible matrices.For reflexive: Since $A = I A I^{-1}$ w

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an equivalence relation

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(D) Let the relation be defined as $R = \{(A, B) : B = P A Q^{-1}\}$ where $P$ and $Q$ are invertible matrices.For reflexive: Since $A = I A I^{-1}$ where $I$ is the identity matrix (which is invertible),$(A, A) \in R$. Thus,$R$ is reflexive.For symmetric: Let $(A, B) \in R$. Then $B = P A Q^{-1}$. Multiplying by $P^{-1}$ on the left and $Q$ on the right,we get $P^{-1} B Q = A$. Since $P^{-1}$ and $Q$ are invertible,let $P' = P^{-1}$ and $Q' = Q^{-1}$. Then $A = P' B (Q')^{-1}$,so $(B, A) \in R$. Thus,$R$ is symmetric.For transitive: Let $(A, B) \in R$ and $(B, C) \in R$. Then $A = P_1 B Q_1^{-1}$ and $B = P_2 C Q_2^{-1}$. Substituting $B$,we get $A = P_1 (P_2 C Q_2^{-1}) Q_1^{-1} = (P_1 P_2) C (Q_1 Q_2)^{-1}$. Since the product of invertible matrices is invertible,$(A, C) \in R$. Thus,$R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

We define a binary relation $\sim$ on the set of all $3 \times 3$ real matrices as $A \sim B$ if and only if there exist invertible matrices $P$ and $Q$ such that $B = P A Q^{-1}$. The binary relation $\sim$ is

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