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 \[PA{P^{ - 1}} = B\]”. Then which of the following true?
A. \[R\] is reflexive, symmetric but not transitive
B. \[R\] is symmetric, transitive but not reflexive
C. \[R\] is an equivalent relation
D. \[R\] is reflexive, transitive but not symmetric

Answer 1

Hint: We will check the relation \[PA{P^{ - 1}} = A\] to check the reflexive relation. Then check whether \[PA{P^{ - 1}} = B\] implies \[PB{P^{ - 1}} = A\] to check whether the relation is symmetric. Then check whether \[PA{P^{ - 1}} = B\] and \[PB{P^{ - 1}} = C\] to check whether the relation is transitive.

Formula used:
If \[ARA\], then relation is reflexive.
If \[ARB\] implies \[BRA\], then relation is symmetric.
If \[ARB\] and \[BRC\] implies \[ARC\], then relation is transitive.
\[AI = A\] where \[I\] is an identity matrix.

Complete step by step solution:
Given that 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 \[PA{P^{ - 1}} = B\]”.
For reflexive:
\[PA{P^{ - 1}} = A\] if \[P = I\] where \[I\] is an identity matrix.
So, \[ARA\] implies \[R\] is reflexive.

For symmetric:
Let \[PA{P^{ - 1}} = B\].
Now pre multiply \[{P^{ - 1}}\] on both sides
\[{P^{ - 1}}PA{P^{ - 1}} = {P^{ - 1}}B\]
Now post multiply \[P\] on both sides
\[\left( {{P^{ - 1}}P} \right)A\left( {{P^{ - 1}}P} \right) = {P^{ - 1}}BP\]
Apply the formula \[A{A^{ - 1}} = I\]
\[IAI = {P^{ - 1}}BP\]
Now applying the formula \[AI = A\]
\[A = {P^{ - 1}}BP\]
Let \[{P^{ - 1}} = M\]. So, \[P = {M^{ - 1}}\]
Now putting \[{P^{ - 1}} = M\] and \[P = {M^{ - 1}}\]
\[A = MB{M^{ - 1}}\]
Hence \[ARB\] implies \[BRA\]. So, \[R\] is symmetric.

For transitive:
Let \[ARB\] and \[BRC\].
\[ARB\] implies \[PA{P^{ - 1}} = B\].
\[BRC\] implies \[PB{P^{ - 1}} = C\].
Now putting \[B = PA{P^{ - 1}}\] in \[PB{P^{ - 1}} = C\].
\[P\left( {PA{P^{ - 1}}} \right){P^{ - 1}} = C\]
\[ \Rightarrow \left( {PP} \right)A\left( {{P^{ - 1}}{P^{ - 1}}} \right) = C\]
\[ \Rightarrow {P^2}A{\left( {{P^{ - 1}}} \right)^2} = C\]
\[ \Rightarrow {P^2}A{\left( {{P^2}} \right)^{ - 1}} = C\]
So, \[ARC\] for matrix \[{P^2}\].
Thus \[R\] is transitive.
Since \[R\] is reflexive, symmetric and transitive. So \[R\] is equivalent relation.

Hence option C is correct.

Note: Many students do a common mistake to check transitive and symmetric relation. Since they get \[A = {P^{ - 1}}BP\] which not same as \[PA{P^{ - 1}} = B\]. So that they thought the relation is not symmetric. Similarly, they conclude that they relation is not transitive. But the relation is symmetric and transitive.