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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
Using properties of determinants, prove that:
$\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
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Answer

$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_{3 }\rightarrow R_{3 }- R_1,$ we have:
$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta-\alpha&\beta^2-\alpha^2&\alpha-\beta\\\gamma-\alpha&\gamma^2-\alpha^2&\alpha-\gamma\end{vmatrix}$
$=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\0&\gamma+\alpha&-1\end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2,$ we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\1&\gamma-\beta&0\end{vmatrix}$
Expanding along $R_3,$ we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$
$=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$
$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$
Hence, the given result is proved.

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