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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
If $\text{a}+\text{b}+\text{c}\neq0\text{ and}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}=0,$ then prove that $a = b = c$.
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Answer

Let $\Delta=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$ $=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_1\big]$
$\Delta=(\text{a}+\text{b}+\text{c})\begin{vmatrix}0&0&1\\\text{b}-\text{a}&\text{c}-\text{a}&\text{a}\\\text{c}-\text{b}&\text{a}-\text{b}&\text{b}\end{vmatrix}$
[Expanding along $R_1$​​​​​​​]
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{a})(\text{a}-\text{b})-(\text{c}-\text{a})(\text{c}-\text{b})]$
$=(\text{a}+\text{b}+\text{c})(\text{ba}-\text{b}^2-\text{a}^2+\text{ab}-\text{c}^2+\text{cb}+\text{ac}-\text{ab})$
$=-(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[2\text{a}^2+2\text{b}^2-2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big]$
$=-\frac{1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}^2+\text{b}^2-2\text{ab})+(\text{b}^2+\text{c}^2-2\text{bc})+(\text{c}^2+\text{a}^2-2\text{ac})\big]$
$-\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]$
Given $\Delta=0$
$\Rightarrow\ \frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]=0$
$\Rightarrow\ (\text{a}-\text{b})^2+(\text{b}-\text{c}^2)+(\text{c}-\text{a})^2=0$ $[\because\ \text{a}+\text{b}+\text{c}\neq0,\text{ given}]$
$\Rightarrow\ \text{a}-\text{b}=\text{b}-\text{c}=\text{c}-\text{a}=0$
$\Rightarrow\ \text{a}=\text{b}=\text{c}$

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