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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants4 Marks
Question
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
✓

Answer

$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$

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