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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^3\\1&\text{b}&\text{b}^3\\1&\text{c}&\text{c}^3\end{vmatrix}=0,\text{b}\neq\text{c}$
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Answer

$\Rightarrow\begin{vmatrix}1&\text{x}&\text{x}^3\\1&\text{b}-\text{x}&\text{b}^3-\text{x}^3\\1&\text{c}-\text{x}&\text{c}^3-\text{x}^3\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2+\text{x}^2+\text{cx}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2+\text{x}^2+\text{cx}-(\text{b}^2+\text{x}^2+\text{bx})\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2-\text{b}^2+\text{cx}-\text{bx}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})(\text{c}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&0&\text{b}+\text{c}+\text{x}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})(\text{c}-\text{b})(\text{b}+\text{c}+\text{x})=0$
$\Rightarrow(\text{b}-\text{x})=0,(\text{c}-\text{x})=0,(\text{b}+\text{c}+\text{x})=0$
$\Rightarrow\text{x}=\text{b},\text{x}=\text{c},\text{x}=-(\text{b}+\text{c})$

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