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 $PA{P}^{-1}=B^{\prime \prime }$. 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 $P(\det (P)\ne 0$ ) such that $PA{P}^{-1}=B$
For reflexive
$ARA\Rightarrow PA{P}^{-1}=A\dots (1)$ must be true
for $P=I,Eq.(1)$ is true so $R^{\prime }$ is reflexive
For symmetric
$ARB\iff PA{P}^{-1}=B\dots (1)$ is true
for BRA iff $PB{P}^{-1}=A\dots (2)$ must be true
$\because PA{P}^{-1}=B$
$P^{-1}PA{P}^{-1}=P^{-1}B$
$IA{P}^{-1}P=P^{-1}BP$
$A=P^{-1}BP\dots (3)$
from $(2)\&(3)PB{P}^{-1}=P^{-1}BP$
can be true some $P=P^{-1}\Rightarrow P^2=I(\det (P)\ne 0)$
So 'R' is symmetric
For trnasitive
$ARB\iff PA{P}^{-1}=B\dots$ is true
$BRC\iff PB{P}^{-1}=C\dots$ is true
now $PPA{P}^{-1}{P}^{-1}=C$
$P^{2}A{\left(P^2\right)}^{-1}=C\Rightarrow ARC$
So 'R' is transitive relation
$\Rightarrow$ Hence $R$ is equivalence