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Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}=0,\text{a}\neq\text{b}$
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Answer

Let $\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$ [Applying $R_2 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\0&\text{x}-\text{b}&\text{x}^2-\text{b}^2\end{vmatrix}$ [Applying $R_3 \rightarrow R_1 - R_3]$
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$

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