#### Provide solution for RD Sharma maths Class 12 Chapter Relations Exercise 1.1 Question 3 Subquestion (iii) maths textbook solution.

$R_{3}$ is reflexive but neither symmetric nor transitive

Hint :

A relation R on set A is

Reflexive relation:

If $(a, a) \in R$ for every $a \in R$

Symmetric relation:

If $(a,b)$ is true then $(b,a)$  is also true for every $a, b \in A$

Transitive relation:

if $(a,b)$ and $(b, c) \in R$, then $(a, c) \in R$  for every  $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{A}$

Given :  $R_{3}$ on R defined by $(a, b) \in R_{3} \Leftrightarrow a^{2}-4 a b+3 b^{2}=0$

Solution :

Reflexivity:

Let a be an arbitrary element of $R_{3}$

Then, $a \in R_{3}$

$a^{2}-4 a \times a+3 a^{2}=0$

So,  $R_{3}$ is reflexive

Symmetry :

Let : $(a, b) \in R_{3}$

$a^{2}-4 a b+3 b^{2}=0$

But  $b^{2}-4 b a+3 a^{2} \neq 0$  for all $a, b \in R$

So, $R_{3}$ is not symmetric.

Transitivity:

Let  $(a, b) \in R_{3} \text { and }(b, c) \in R_{3}$

\begin{aligned} &a^{2}-4 a b+3 b^{2}=0\; \; \; \; \; \; ...(i)\\ &\text { and } b^{2}-4 b c+3 c^{2}=0 \; \; \; \; \; \; \; ....(ii) \end{aligned}

Adding (i) and (ii) we get,

\begin{aligned} &\qquad a^{2}-4 a b+3 b^{2}+b^{2}-4 b c+3 c^{2}=0 \\ &\Rightarrow a^{2}-4 a b+4 b^{2}-4 b c+3 c^{2}=0 \\ &\text { But } a^{2}-4 a c+3 c^{2}=-4 a c-4 a b+4 b^{2}-4 b c \neq 0 \\ &\Rightarrow a^{2}-4 a c+3 c^{2} \neq 0 \end{aligned}

$(a, c) \notin R_{3}$

So, $R_{3}$ is not transitive.