JEE PYQ: Sets And Relations Question 3
Question 3 - 2021 (18 Mar Shift 2)
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?
(1) R is symmetric, transitive but not reflexive
(2) R is reflexive, symmetric but not transitive
(3) R is an equivalence relation
(4) R is reflexive, transitive but not symmetric
Show Answer
Answer: (3)
Solution
A and $B$ are matrices of $n \times n$ order & ARB iff there exists a non-singular matrix P ($\det(P) \neq 0$) such that $PAP^{-1} = B$.
For reflexive: $ARA \Rightarrow PAP^{-1} = A$ … (1) must be true.
For $P = I$, Eq.(1) is true so ‘R’ is reflexive.
For symmetric: $ARB \Leftrightarrow PAP^{-1} = B$ … (1) is true.
For BRA iff $PBP^{-1} = A$ … (2) must be true. Since $PAP^{-1} = B$, we get $P^{-1}PAP^{-1}P = P^{-1}BP$, so $A = P^{-1}BP$.
From (2) & (3): $PBP^{-1} = P^{-1}BP$ can be true for some $P = P^{-1} \Rightarrow P^2 = I$ ($\det(P) \neq 0$).
So ‘R’ is symmetric.
For transitive: $ARB \Leftrightarrow PAP^{-1} = B$ … is true, $BRC \Leftrightarrow PBP^{-1} = C$ … is true.
Now $P^2A(P^2)^{-1} = C \Rightarrow ARC$.
So ‘R’ is transitive relation.
$\Rightarrow$ Hence R is equivalence.
BETA
