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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
If $a, b, c$ are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either $a + b + c = 0$ or $a = b= c.$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix} [$Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}-\text{a}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix} [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=2(\text{a}+\text{b}+\text{c})\left\{1\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{a}\\\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}\right\}$
$=2(\text{a}+\text{b}+\text{c})\{(\text{b}-\text{c})(\text{c}-\text{b})-(\text{b}-\text{a})(\text{c}-\text{a})\}$
$=-2(\text{a}+\text{b}+\text{c})\{\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}$
But $\triangle=0 [$Given$]$
$\Rightarrow-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}=0$
Either,
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2=0$
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $\text{a}=\text{b}=\text{c}$
Hence proved.

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