Verified Solution • 2025–26 Session moderate

QJEEM21Ma94a9: JEE Main 2021 Mathematics - Class 11 JEEM21Ma94a9 Solution: Define a relation R over a class of n × n real matrices A and B …

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 the following is true?

Multiple Choice Options

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

Step-by-Step Verified Solution

For reflexive relation,

∀(A,A)∈R\forall(A, A) \in R for matrix PP .

⇒A=PAP−1\Rightarrow A=P A P^{-1} is true for P=1P=1

So, RR is reflexive relation.

For symmetric relation,

Let (A,B)∈R(A, B) \in R for matrix PP .

⇒A=PBP−1

After pre-multiply by P−1P^{-1} and post-multiply by PP , we get

P−1AP=B

So, (B,A)∈R(B, A) \in R for matrix P−1P^{-1} .

So, RR is a symmetric relation.

For transitive relation,

Let ARBA R B and BRCB R C

So, A=PBP−1A=P B P^{-1} and B=PCP−1B=P C P^{-1}

Now, A=P(PCP−1)P−1A=P\left(P C P^{-1}\right) P^{-1}

⇒A=(P)2C(P−1)2⇒A=(P)2⋅C⋅(P2)−1\Rightarrow A=(P)^2 C\left(P^{-1}\right)^2 \Rightarrow A=(P)^2 \cdot C \cdot\left(P^2\right)^{-1}

∴(A,C)∈R\therefore(A, C) \in R for matrix P2P^2 .

∴R\therefore R is transitive relation.

Hence, RR is an equivalence relation.

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