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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using properties of determinants, prove that,
$\begin{vmatrix}1& 1&1+3\text{x} \\1+3\text{y} & 1&1\\1&1+3\text{z}&1 \end{vmatrix}=9(3\text{xyz}+\text{xy}+\text{yz}+\text{zx})$
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Answer

$\begin{vmatrix}1& 1&1+3\text{x} \\1+3\text{y} & 1&1\\1&1+3\text{z}&1 \end{vmatrix}$
$\text{xyz}\begin{vmatrix}\frac{1}{\text{x}}& \frac{1}{\text{x}}&\frac{1}{\text{x}}+3 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3$
$=\text{xyz}\begin{vmatrix}\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3&\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3&\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$=(\text{xyz})\bigg(\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3\bigg)\begin{vmatrix}1 & 1&1 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$\text{C}_2\rightarrow\text{C}_2-\text{C}_1\ \&\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
$=\text{xyz}\bigg(\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3\bigg)\begin{vmatrix}1 & 0&0 \\\frac{1}{\text{y}}+3 & -3&-3\\\frac{1}{\text{z}}&3&0 \end{vmatrix}$
$=(\text{yz}+\text{zx}+\text{xy}+3\text{xyz})(0+9)$
$=9(3\text{xyz}+\text{xy}+\text{yz}+\text{zx})=\text{R.H.S.}$

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