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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Using properties of determinants, prove that:
$\begin{vmatrix} \text{(b + c)}^{2} & \text{a}^{2} & \text{bc} \\ \text{(c + a)}^{2} & \text{b}^{2} & \text{ca} \\ \text{(a + b)}^{2} & \text{c}^{2} & \text{ab} \end{vmatrix} = {(a - b) (b - c) (c - a) (a + b + c)}\text{(a}^{2} + \text{b}^{2} + \text{c}^{2}) $
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Answer

$\text{Let}\Delta = \begin{vmatrix} (b + c)^{2} & a^{2} & bc \\ (c + a)^{2} & b^{2} & ca \\ (a + b)^{2} & c^{2} & ab \end{vmatrix}$
$\text{C}_{1}\rightarrow\text{C}_{1} + \text{C}_{2} - \text{2C}_{3} \Rightarrow \Delta = (a^{2} + b^{2} + c^{2}) \begin{vmatrix} 1 & a^{2} & bc \\ 1 & b^{2} & ca \\ 1 & c^{2} & ab \end{vmatrix}$
$\text{R}_{1}\rightarrow\text{R}_{1} + \text{R}_{2}, \text{and } \text{R}_{2} \rightarrow\text{R}_{2} - \text{R}_{3} \Rightarrow \Delta = (a^{2} + b^{2} + c^{2}) \begin{vmatrix} 0 & a^{2} - b^{2} & c(b - a) \\ 0 & b^{2} - c^{2} & a(c - b) \\ 1 & c^{2} & ab \end{vmatrix} $
$= (a^{2} + b^{2} + c^{2})(a - b)(b - c) \begin{vmatrix} 0 & a + b & -c \\ 0 & b + c & -a \\ 1 & c^{2} & ab \end{vmatrix} $
$\text{R}_{1}\rightarrow\text{R}_{2} - \text{R}_{1}\Rightarrow\Delta= (a^{2} + b^{2} + c^{2})(a - b)(b - c) \begin{vmatrix} 0 & a + b & -c \\ 0 & c -a & c -a\\ 1 & c^{2} & ab \end{vmatrix} $
$\therefore\Delta= (a^{2} + b^{2} + c^{2})(a - b)(b - c)(c - a) \begin{vmatrix} 0 & a + b & -c \\ 0 & 1 & 1 \\ 1 & c^{2} & ab \end{vmatrix} $
$\text{Expanding by C}_{1} \text{ to get } \Delta = (a^{2} + b^{2} + c^{2})(a - b)(b - c)(c - a)(a + b + c) $

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