$A$ and $B$ are matrices of $n \times n$ order  ARB iff

there exists a non singular matrix $P (\operatorname{det}( P ) \neq 0)$

such that $PAP ^{-1}= B$

For reflexive

$ARA \Rightarrow PAP ^{-1}= A \quad \ldots(1)$ must be true for $P = I ,$ Eq.(1) is true so $'R'$ is reflexive

For symmetric

$ARB \Leftrightarrow PAP ^{-1}= B \quad \ldots(1)$ is true

for $BRA$ iff $PBP ^{-1}= A \ldots$. (2) must be true

$\because PAP -1= B$

$P ^{-1} PAP ^{-1}= P ^{-1} B$

$IAP ^{-1} P = P ^{-1} BP$

$A = P ^{-1} BP \ldots(3)$

from $(2)$ And $(3) PBP ^{-1}= P ^{-1} BP$

can be true some $P = P ^{-1} \Rightarrow P ^{2}= I (\operatorname{det}( P ) \neq 0)$

So $'R'$ is symmetric

For trnasitive

$ARB \Leftrightarrow PAP ^{-1}= B \ldots$ is true

$BRC \Leftrightarrow PBP ^{-1}= C \ldots$ is true

now $\quad PPAP ^{-1} P ^{-1}= C$

$P ^{2} A \left( P ^{2}\right)^{-1}= C \Rightarrow ARC$

So $'R'$ is transitive relation

$\Rightarrow$ Hence $R$ is equivalence