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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 16 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 16 Maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)

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

Given:\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)

Solution:

\text { L.H.S }\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|

If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.

=\left|\begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2} \end{array}\right|

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

\begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ b^{2}-a^{2} & c^{2}-b^{2} & a^{2}+b^{2} \end{array}\right| \\ &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ (b-a)(b+a) & (c-b)(c+b) & a^{2}+b^{2} \end{array}\right| \quad \because x^{2}-y^{2}=(x-y)(x+y) \end{aligned}

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

\begin{aligned} &\text { On expanding w.r.t } R_{1}\\ &=(b-a)(c-b)\left[0\left|\begin{array}{cc} 1 & a+b \\ c+b & a^{2}+b^{2} \end{array}\right|-0\left|\begin{array}{cc} 1 & a+b \\ b+a & a^{2}+b^{2} \end{array}\right|+1\left|\begin{array}{cc} 1 & 1 \\ b+a & c+b \end{array}\right|\right]\\ &=(-)(a-b)(-)(b-c)[0-0+1\{(c+b) 1-1(b+a)\}]\\ &=(a-b)(b-c)[1\{(c+b-b-a)\}]\\ &=(a-b)(b-c)(c-a)\\ &=R H . S \end{aligned}

Hence it is proved that

\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)

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