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Determinants — Maths STD 12 Science — Question

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

${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 x2 - y2 = (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 x2 - y2 = (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]
$=-(a-b)(b-c)(a+b+c)(c-a)\{-2c^2+2{ab}-a^2-b^2-2{ab}+c^2\}$
$=-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2)$ $=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$ $={R.H.S}$

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