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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}$
$=\begin{vmatrix}0&(\text{b}+\text{c})-(\text{c}+\text{a})&(\text{b}^2+\text{c}^2)-(\text{c}^2+\text{a}^2)\\0&(\text{c}+\text{a})-(\text{a}+\text{b})&(\text{c}^2+\text{a}^2)-(\text{a}^2+\text{b}^2)\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Applying $R_1 → R_1 - R_2$ and $R_2 → R_2 - R_3]$
$=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}^2-\text{a}^2\\0&\text{c}-\text{b}&\text{c}^2-\text{b}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(-1)^2\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{b}^2\\0&\text{b}-\text{c}&\text{b}^2-\text{c}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Taking out $(-1)$ common from $R_1$ and $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&\text{a}+\text{b}\\0&1&\text{b}+\text{c}\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(\text{a}+\text{b})(\text{b}-\text{c})\left\{1\times\begin{vmatrix} 1 & \text{a}+\text{b} \\ 1 & \text{b}+\text{c} \end{vmatrix}\right\}$
$=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
$=\text{R.H.S}$

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