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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}$
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Answer

$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{a}+\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R3 → R3 + R2]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\1&1&1\end{vmatrix}$ [Taking (a + b + c) common]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\1&1&1\end{vmatrix}$ [Applying R2 → R1 - R2]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}-\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}-\text{a}&\text{a}\\0&0&1\end{vmatrix}$ [C1 → C1 - C2 and C2 → C2 - C3]
$=(\text{a}+\text{b}+\text{c})\big[-1\{(\text{a}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{c})^2\}\big]$
$=(\text{a}+\text{b}+\text{c})\big[-\{\text{ac}-\text{bc}-\text{a}^2+\text{ab}-\text{b}^2-\text{c}^2+2\text{bc}\}\big]$
$=(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}=\text{ R.H.S}$

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