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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
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Answer

$\triangle=\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{bc}-\text{b}^2+\text{ac}\\0&\text{b}-\text{c}&\text{b}^2-\text{ac}-\text{c}^2+\text{ab}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$ [applying $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3$]
$=\begin{vmatrix}0&\text{a}-\text{b}&(\text{a}-\text{b})(\text{a}+\text{b})+\text{c}(\text{a}-\text{b})\\0&\text{b}-\text{c}&(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&(\text{a}+\text{b}+\text{c})\\0&1&(\text{a}+\text{b}+\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$\Rightarrow\triangle=0$

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