Question

Define a relation $\mathrm{R}$ over a class of $\mathrm{n}\times \mathrm{n}$ real matrices $\mathrm{A}\&\mathrm{B}$ as “$\mathrm{ARB}$ if there exists a non-singular matrix $\mathrm{P}$ such that $PA{P}^{-1}=B$”.

Then which of the following is true?

• A. $\mathrm{R}$ is reflexive, symmetric but not transitive
• B. $\mathrm{R}$ is symmetric, transitive but not reflexive
• C. $\mathrm{R}$ is an equivalence relation
• D. $\mathrm{R}$ is reflexive, Transitive but not symmetric

Answer 1

The correct option is C

$\mathrm{R}$ is an equivalence relation

Explanation for correct option.

Step1. For reflexive

Given, a relation $\mathrm{R}$ over a class of $\mathrm{n}\times \mathrm{n}$ real matrices $\mathrm{A}\&\mathrm{B}$ as “$\mathrm{ARB}$ if there exists a non-singular matrix $\mathrm{P}$ such that $PA{P}^{-1}=B$”

$$\begin{gathered} (B,B)\in R \\ B=PB{P}^{-1} \end{gathered}$$

Which is true for $\mathrm{P}=\mathrm{I}$

‘$\mathrm{R}$ is reflexive.

Step2. For symmetric.

As $(B,A)\in R$for matrix $\mathrm{P}$

$\begin{array}{lcl}& & B=PA{P}^{-1}\\ \Rightarrow P^{-1}B& =& P^{-1}PA{P}^{-1}\end{array}$

$\begin{array}{lcl}& & P^{-1}BP\\ & =& IA{P}^{-1}P\\ & =& IAI\end{array}$

$\begin{array}{rcl}\Rightarrow P^{-1}BP& =& A\\ & & A=P^{-1}BP\end{array}$

$(A,B)\in R$ for matrix $P^{-1}$

$\mathrm{R}$ is symmetric

Step3. For transitive.

$B=PA{P}^{-1}$

$\mathrm{A}={\mathrm{PCP}}^{-1}$

Now,

$\begin{array}{rcl}B& =& P\left(PC{P}^{-1}\right)P^{-1}\\ B& =& P^{2}C{\left(P^{-1}\right)}^2\\ & & B=P^{2}C{\left(P^2\right)}^{-1}\end{array}$

$(B,C)\in R$ for matrix $P^2$

$\mathrm{R}$ is transitive

So, $\mathrm{R}$ is an equivalence relation.

Hence, correct option is$\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{.}$