Question

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 ?

• A. $R$ is symmetric, transitive but not reflexive,
• B. $R$ is reflexive, symmetric but not transitive
• C. $R$ is an equivalence relation
• D. $R$ is reflexive, transitive but not symmetric

Answer 1

Answer: (C)

$A$ 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
$ARA \Rightarrow PAP ^{-1}= A \ldots(1)$ must be true
for $P = I , Eq .(1)$ is true so $R ^{\prime}$ is reflexive
For symmetric
$ARB \Leftrightarrow PAP ^{-1}= B \ldots(1)$ 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 $(2) \&(3) PBP ^{-1}= P ^{-1} BP$
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 $PPAP ^{-1} P ^{-1}= C$
$P ^{2} A \left( P ^{2}\right)^{-1}= C \Rightarrow ARC$
So 'R' is transitive relation
$\Rightarrow $ Hence $R$ is equivalence