Define a relation

Solution:

$\mathrm{A}$ and $B$ are matrices of $n \times n$ order \& ARB iff there exists a non singular matrix $\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0)$ such that $\mathrm{PAP}^{-1}=\mathrm{B}$

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

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

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

$\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}$

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

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

from (2) $\backslash$ (3) $\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}$

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

So ' $R$ ' is symmetric For trnasitive

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

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

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

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

So 'R' is transitive relation

$\Rightarrow$ Hence $\mathrm{R}$ is equivalence