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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?
$R$ is symmetric, transitive but not reflexive.
$R$ is reflexive, symmetric but not transitive
$R$ is an equivalence relation
$R$ is reflexive, transitive but not symmetric
Solution
$A$ 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