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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
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Answer

Let $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
[Taking $x^2$ common from from $C_1, y^2$ common from $C_2$ and $z^2$ common from $C_3$]
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
[Applying $C_2 → C_2 - C_3​​​​​​​$]
$=\text{x}^3\text{y}^3\text{z}^3(1+1)$ [Expanding along first row]
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$

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