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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{y}^2\text{z}^2&\text{yz}&\text{y}+\text{z}\\\text{z}^2\text{x}^2&\text{zx}&\text{z}+\text{x}\\\text{x}^2\text{y}^2&\text{xy}&\text{x}+\text{y}\end{vmatrix}=0$
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Answer

We have, $\begin{vmatrix}\text{y}^2\text{z}^2&\text{yz}&\text{y}+\text{z}\\\text{z}^2\text{x}^2&\text{zx}&\text{z}+\text{x}\\\text{x}^2\text{y}^2&\text{xy}&\text{x}+\text{y}\end{vmatrix}$
$[\text{Multiplying R}_1, \text{R}_2, \text{R}_3\text{ by x, y, z respectively}]$
$=\frac{1}{\text{xyz}}\begin{vmatrix}\text{xy}^2\text{z}^2&\text{xyz}&\text{xy}+\text{xz}\\\text{x}^2\text{yz}^2&\text{xyz}&\text{yz}+\text{xy}\\\text{x}^2\text{y}^2\text{z}&\text{xyz}&\text{xz}+\text{yz}\end{vmatrix}$
$\big[\text{Taking (xyz) common from C}_1\text{ and C}_2\big]$
$=\frac{1}{\text{xyz}}(\text{xyz})^2\begin{vmatrix}\text{yz}&1&\text{xy}+\text{xz}\\\text{xz}&1&\text{yz}+\text{xy}\\\text{xy}&1&\text{xz}+\text{yz}\end{vmatrix}$
$[\text{Applying C}_3\rightarrow\text{C}_3+\text{C}_1]$
$=\text{xyz}\begin{bmatrix}\text{yz}&1&\text{xy}+\text{yz}+\text{zx}\\\text{xz}&1&\text{xy}+\text{yz}+\text{zx}\\\text{xy}&1&\text{xy}+\text{yz}+\text{zx}\end{bmatrix}$
$\big[\text{Taking (xy}+\text{yz}+\text{zx})\text{ common from }\text{C}_3\big]$
$=\text{xyz (yz}+\text{yz}+\text{zx})\begin{vmatrix}\text{yz}&1&1\\\text{xz}&1&1\\\text{xy}&1&1\end{vmatrix}$
$=0$
$\big[\because\text{C}_2\text{ and C}_3\text{ are identical}\big]$

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