$(A, A) \in R$

$A=P^{-1} A P$ is true,

For $P=I,$ which is an invertible matrix.

$\therefore \mathrm{R}$ is reflexive.

For symmetry

As $(A, B) \in R$ for matrix $P$

$\boldsymbol{A}=\boldsymbol{P}^{-1} \boldsymbol{B} \boldsymbol{P}$

$\Rightarrow P A P^{-1}=B$

$\Rightarrow B=P A P^{-1}$

$\Rightarrow B=\left(P^{-1}\right)^{-1} A\left(P^{-1}\right)$

$\therefore(B, A) \in$ for matrix $R P^{-1}$

$\therefore R$ is symmetric

For transitivity

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

and $B=P^{-1} C P$

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

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

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

$\therefore(A, C) \in R$ for matrix $P^{2}$

$\therefore R$ is transitive.

So $R$ is equivalence