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  • Need Solution for R D Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 11 Maths Textbook Solution.

Need Solution for R D  Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 2 sub question 11 Maths Textbook Solution.

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Answer:\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c

Hint: We will try to make some elements of the determinant zero

Given:\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|

Solution: L.H.S\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|

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

\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{1}\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 0 & b-c & c-a \\ 2 & c+a & a+b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding wr.t } \mathrm{R}_{1}\\ &=(a+b+c)\left[1\left|\begin{array}{ll} b-c & c-a \\ c+a & a+b \end{array}\right|-b\left|\begin{array}{ll} 0 & c-a \\ 2 & a+b \end{array}\right|+c \mid \begin{array}{ll} 0 & b-c \\ 2 & c+a \end{array}\right]\\ &=(a+b+c)[1\{(b-c)(a+b)-(c-a)(c+a)\}-b\{0 \times(a+b)-2(c-a)\}+c\{0 \times(c+a)-2(b-c)\}]\\ &=(a+b+c)\left[1\left\{\left(a b-c a+b^{2}-b c\right)-\left(c^{2}-a^{2}\right)\right\}-b\{0-2 c+2 a\}+c\{0-2 b+2 c\}\right]\\ &=(a+b+c)\left[a b-c a+b^{2}-b c-c^{2}+a^{2}+2 b c-2 a b-2 b c+2 c^{2}\right] \end{aligned}

\begin{aligned} &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ &=a^{3}+b^{3}+c^{3}-3 a b c \quad \because x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \end{aligned}

=RHS

Hence it is proved that

\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c

 

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