Question

We define a binary relation ~ on the set of all 3x3 matrices denoted as ${{M}_{3\times 3}}$ as A~B if and only if there exist invertible matrices P and Q such that $B=PA{{Q}^{-1}}$. The relation is
[a] Neither reflexive nor symmetric
[b] Reflexive and symmetric but not transitive
[c] Symmetric and transitive but not reflexive
[d] An equivalence relation

Answer 1

Hint: Use the fact that every matrix can be written as $A=IA{{I}^{-1}}$ and hence prove that the relation is reflexive. Use the fact that if $A=PB{{Q}^{-1}}$ then $B={{P}^{-1}}AQ$ and hence prove that the given relation is reflexive. Use the fact that if $PA{{Q}^{-1}}=RC{{T}^{-1}}$, then $A={{P}^{-1}}RC{{T}^{-1}}Q$ and hence prove that the given relation is transitive. Use the fact that if a relation is symmetric reflexive and transitive then the relation is an equivalence relation. Hence determine which of the options is correct.

Complete step-by-step answer:
Before solving the question, we need to understand what reflexive, symmetric and transitive relations are
Reflexive relation: A relation R on set A is said to be reflexive if $\forall a\in A$, $\left( a,a \right)\in R$. This means every element of A is related to itself, then R is reflexive.
Symmetric relation: A relation R on set A is said to be reflexive if $aRb\Rightarrow bRa$ i.e. both pairs (a,b) and (b,a) are in R.
Transitive relation: A relation R on set A is said to be transitive if $aRb,bRc\Rightarrow aRc$.
Equivalence relation: A relation, which is reflexive, symmetric and transitive is called an equivalence relation.
Reflexivity:
We know that $A=IA=AI$
Hence, we have
$A=IAI$
Since $I={{I}^{-1}}$, we have
$A=IA{{I}^{-1}}$
Since $I$ is invertible, we have
$A\sim A$
Hence, we have ~ is Reflexive.
Symmetricity:
We have A~B.
Hence, we have $B=PA{{Q}^{-1}}$, where P and Q are invertible. Since $P$ is invertible, pre-multiplying both sides by ${{P}^{-1}}$, we get
${{P}^{-1}}B={{P}^{-1}}PA{{Q}^{-1}}$
We know that ${{P}^{-1}}P=I$
Hence, we have
${{P}^{-1}}B=IA{{Q}^{-1}}=A{{Q}^{-1}}$
Post-multiplying both sides by Q, we get
${{P}^{-1}}BQ=A{{Q}^{-1}}Q=AI=A$
Hence, we have $A={{P}^{-1}}B{{\left( {{Q}^{-1}} \right)}^{-1}}$
Since P and Q are invertible, we have ${{P}^{-1}}$ and ${{Q}^{-1}}$ are also invertible.
Hence, we have B~A.
Hence the relation is symmetric.
Transitivity:
We have A~B and B~C.
Since A~B, we have $B=PA{{Q}^{-1}}$ where P and Q are invertible.
Since B~C, we have $C=RB{{T}^{-1}}$, where R and T are invertible.
Pre-multiplying both sides by ${{R}^{-1}}$, we get
${{R}^{-1}}C=B{{T}^{-1}}$
Post-multiplying both sides by T, we get
${{R}^{-1}}CT=B$
Hence, we have
${{R}^{-1}}CT=PA{{Q}^{-1}}$
Pre-multiplying both sides by R, we get
$CT=RPA{{Q}^{-1}}$
Post-multiplying both sides by ${{T}^{-1}}$, we get
$C=RPA{{Q}^{-1}}{{T}^{-1}}$
We know that ${{B}^{-1}}{{A}^{-1}}={{\left( AB \right)}^{-1}}$
Hence, we have
$C=RPA{{\left( TQ \right)}^{-1}}$
Since R and P are invertible, we have RP is invertible.
Similarly, since T and Q are invertible, we have TQ is invertible,
Hence, we have
A~C
Hence, the relation is transitive.
Since the relation is reflexive, symmetric and transitive, we have the relation as an equivalence relation.
Hence option [d] is correct.

Note: [1] In these types of questions students make mistakes in checking reflexivity of the relation. They show that a is related to a for some elements in A and claim that the relation is reflexive, which is incorrect since, we need to show it is true for every element in A.