Define a relation R over a class of n× n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP^-1=B". Then which of…

Given $A$A and $B$B are matrices of $n\times n$n×n order and $ARB$ARB iff there exists a non-singular matrix $P(det(P)\neq 0)$P(det(P)=0) such that $PA{P}^{-1}=B$PAP−1=B.

Reflexivity Check :

$ARA\Rightarrow PA{P}^{-1}=A$ARA⇒PAP−1=A which is true for $P=I$P=I.

So, $R$R is reflexive relation.

Symmetric Check :

$ARB\Rightarrow PA{P}^{-1}=B\Rightarrow {P}^{-1}PA{P}^{-1}P={P}^{-1}BP\Rightarrow IAI={P}^{-1}BP\Rightarrow {P}^{-1}BP=A\Rightarrow BRA$ARB⇒PAP−1=B⇒P−1PAP−1P=P−1BP⇒IAI=P−1BP⇒P−1BP=A⇒BRA for matrix ${P}^{-1}$P−1.

So, $R$R is symmetric relation.

Transitivity Check :

$ARB\Rightarrow PA{P}^{-1}=B$ARB⇒PAP−1=B and $BRC\Rightarrow PB{P}^{-1}=C$BRC⇒PBP−1=C.

So, $PPA{P}^{-1}{P}^{-1}=C$PPAP−1P−1=C.

$$\Rightarrow {P}^{2}A{({P}^{2})}^{-1}=C\Rightarrow ARC$$⇒P2A(P2)−1=C⇒ARC

So, $R$R is a transitive relation.

Since, $R$R is reflexive, symmetric and transitive all.

Hence, $R$R is an equivalence relation.