Define a relation $$R$$ over a class of $$n \times n$$ real matrices $$A$$ and $$B$$ as "$$ARB$$" iff there exists a non-singular matrix $$P$$ such that $$PAP^{-1} = B$$. Then which of the following is true?

We check whether the relation $$ARB \iff PAP^{-1} = B$$ for some non-singular $$P$$ is an equivalence relation.

Reflexive: Taking $$P = I$$ (the identity matrix), we get $$IAI^{-1} = A$$, so $$ARA$$ holds for every matrix $$A$$.

Symmetric: If $$ARB$$, then $$PAP^{-1} = B$$ for some non-singular $$P$$. This gives $$A = P^{-1}BP$$, which means $$QBQ^{-1} = A$$ where $$Q = P^{-1}$$ is also non-singular. So $$BRA$$ holds.

Transitive: If $$ARB$$ and $$BRC$$, then $$PAP^{-1} = B$$ and $$QBQ^{-1} = C$$ for non-singular $$P$$ and $$Q$$. Substituting, $$C = Q(PAP^{-1})Q^{-1} = (QP)A(QP)^{-1}$$. Since $$QP$$ is non-singular, $$ARC$$ holds.

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation.