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  • Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 49 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 49  Maths Textbook Solution.

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Answer: a+b+c=0 \& a=b=c

Hint Use determinant formula

Given:   $$ \text { If } a, b, c \text { are real no. such that }\left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 \text { show that either } a+b=c \text { or } a=b=c

Solution:

$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \\ &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \end{aligned}

$$ \begin{aligned} &2(a+b+c) \text { common from } \mathrm{R}_{1} \\ &=2(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \end{aligned}

$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}-C_{2} \& \mathrm{C}_{2} \rightarrow C_{2}-C_{3} \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ c-b & a-c & b+c \\ a-c & b-a & c+a \end{array}\right| \end{aligned}

$$ \begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=2(a+b+c)\left\{1 \mid \begin{array}{ll} c-b & a-c \\ a-c & b-a \end{array}\right\}\\ &(a+b+c)=0 \quad \text { or } \quad b c-b^{2}-a c+a b-(a-c)^{2}=0\\ &b c-b^{2}-a c+a b=a^{2}+c^{2}-2 a c\\ &a^{2}+b^{2}+c^{2}-a^{2}-a b-a c=0\\ &2\left(a^{2}+b^{2}+c^{2}\right)-2 a c-2 a b-2 b c=0\\ &(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0\\ &a-b=0, \quad b-c=0, \quad c-a=0\\ &a=b, b=c, \quad c=a \end{aligned}

Hence,a+b+c=0 \& a=b=c

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