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

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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 :


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.


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.

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