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#### Explain Solution R.D. Sharma Class 12 Chapter 5.2 deteminants Exercise MCQs Question 2 sub question 3 maths Textbook Solution.

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

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

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

Solution :-$\left|\begin{array}{lll} a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2} \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}\\ &=\left|\begin{array}{lll} a+b+c & b+c & a^{2} \\ b+c+a & c+a & b^{2} \\ c+a+b & a+b & c^{2} \end{array}\right|\\ &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{1}\\ &=a+b+c\left|\begin{array}{lll} 1 & b+c & a^{2} \\ 1 & c+a & b^{2} \\ 1 & a+b & c^{2} \end{array}\right| \end{aligned}

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

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

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

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

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