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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS4 Marks
Question
Prove that:
$\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\$\text{c}+\text{a})^2&\text{b}^2&\text{ca}\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
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Answer

$\text{L.H.S}=\begin{vmatrix}(\text{b}+\text{c})^2&\text{a}^2&\text{bc}\$\text{c}+\text{a})^2&\text{b}^2&\text{ca}\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}(\text{b}+\text{c})^2-(\text{c}+\text{a})^2&\text{a}^2-\text{b}^2&\text{bc}-\text{ca}\$\text{c}+\text{a})^2-(\text{a}+\text{b})^2&\text{b}^2-\text{c}^2&\text{ca}-\text{ab}\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying R1 → R1 - R2 and R2 → R2 - R1]
$=\begin{vmatrix}(\text{b}+\text{c})(\text{b}+2\text{c}+\text{a})&(\text{a}+\text{b})(\text{a}-\text{b})&\text{c}(\text{b}-\text{a})\$\text{c}-\text{a})(\text{b}+2\text{a}+\text{c})&(\text{b}-\text{c})(\text{b}+\text{c})&\text{a}(\text{c}-\text{b})\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-(\text{b}+2\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-(\text{b}+2\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b})^2&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying x2 - y2 = (x + y)(x - y) and taking out (a - b) common from R1 and (b - c) from R2]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b})^2-\text{c}^2&\text{c}^2&\text{ab}\end{vmatrix}$ [Applying C1 → C1 - C2]
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}-2(\text{b}+\text{c}+\text{a})&\text{a}+\text{b}&-\text{c}\\-2(\text{b}+\text{a}+\text{c})&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b}+\text{c})(\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying x2 - y2 = (x + y)(x - y) in C1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\-2&\text{b}+\text{c}&-\text{a}\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Taking out (a + b + c) common from C1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&\text{c}-\text{a}&\text{c}-\text{a}\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Applying R2 → R2 - R1]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}&-\text{c}\\0&1&1\$\text{a}+\text{b}-\text{c})&\text{c}^2&\text{ab}\end{vmatrix}$
[Taking out (c - a) common from R2]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&-\text{c}\\0&0&1\$\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}&\text{ab}\end{vmatrix}$
[Applying C2 → C2 - C3]
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a}) \left\{(-1)\begin{vmatrix}-2&\text{a}+\text{b}+\text{c}&\$\text{a}+\text{b}-\text{c})&\text{c}^2-\text{ab}\end{vmatrix}\right\}$ [Expanding along R2]
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})\{-2\text{c}^2+2\text{ab}-\text{a}^2-\text{b}^2-2\text{ab}+\text{c}^2\}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{a}+\text{b}+\text{c})(\text{c}-\text{a})(-\text{a}^2-\text{b}^2-\text{c}^2)$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$

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