JEE Main 18 March 2021 Shift 2 Solved Paper

For reflexive relation,
∀(A,A)∈R for matrix P.
⇒‌‌A=PAP−1 is true for P=1
So, R is reflexive relation.
For symmetric relation,
Let (A,B)∈R for matrix P.
⇒‌‌A=PBP−1
After pre-multiply by P−1 and post-multiply by Pt
we get
P−1AP=B
So, (B,A)∈R for matrix P−1.
So, R is a symmetric relation.
For transitive relation,
Let ARB and BRC
So, ‌‌A=PBP−1 and B=PCP−1
Now, ‌‌A=P(PCP−1)P−1
⇒‌‌A=(P)2C(P−1)2⇒A=(P)2⋅C⋅(P2)−1
∴(A,C)∈R for matrix P2.
∴R is transitive relation.
Hence, R is an equivalence relation.