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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that: $\begin{vmatrix}(b+c)^2&a^2&\text{bc}\\(c+a)^2&b^2&\text{ca}\\(a+b)^2&c^2&\text{ab}\end{vmatrix}$ $=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$
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Answer

$\text{L.H.S}=\begin{vmatrix}(b+c)^2&a^2&bc\\(c+a)^2&b^2&ca\\(a+b)^2&c^2&ab\end{vmatrix}$
$=\begin{vmatrix}(b+c)^2-(c+a)^2&a^2-b^2&bc-ca\\(c+a)^2-(a+b)^2&b^2-c^2&ca-ab\\(a+b)^2&c^2&ab\end{vmatrix}$ [Applying R1 → R1 - R2 and R2 → R2 - R1]
$=\begin{vmatrix}(b+c)(b+2c+a)&(a+b)(a-b)&c(b-a)\\(c-a)(b+2a+c)&(b-c)(b+c)&a(c-b)\\(a+b)^2&c^2&ab\end{vmatrix}$
$=(a-b)(b-c)\begin{vmatrix}-(b+2c+a)&a+b&-c\\-(b+2a+c)&b+c&-a\\(a+b)^2&c^2&ab\end{vmatrix}$ [Applying $x^2 - y^2 = (x + y)(x - y)$ and taking out $(a - b)$ common from R1 and $(b - c)$ from R2] 
$=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b)^2-c^2&c^2&ab\end{vmatrix}$ [Applying C1 → C1 - C2] $=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b+c)(a+b-c)&c^2&ab\end{vmatrix}$ [Applying $x^2 - y^2 = (x + y)(x - y)$ in C1] $=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\-2&b+c&-a\\(a+b-c)&c^2&ab\end{vmatrix}$ [Taking out $(a + b + c)$ common from C1] $=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\0&c-a&c-a\\(a+b-c)&c^2&ab\end{vmatrix}$ [Applying R2 → R2 - R1] $=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b&-c\\0&1&1\\(a+b-c)&c^2&ab\end{vmatrix}$ [Taking out $(c - a)$ common from R2] $=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b+c&-c\\0&0&1\\(a+b-c)&c^2-ab&ab\end{vmatrix}$ [Applying C2 → C2 - C3] $=(a-b)(b-c)(a+b+c)(c-a) \left\{(-1)\begin{vmatrix}-2&a+b+c&\\(a+b-c)&c^2-ab\end{vmatrix}\right\}$ [Expanding along R2] $=-(a-b)(b-c)(a+b+c)(c-a)\{-2c^2+2ab-a^2-b^2-2ab+c^2\}$
$=-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2)$
$=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$ 
$=\text{R.H.S}$

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