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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
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Answer

We have to prove,
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
$\therefore\ \text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=\begin{vmatrix}\text{y}+\text{z}+\text{z}+\text{y}&\text{z}&\text{y}\\\text{z}+\text{z}+\text{x}+\text{x}&\text{z}+\text{x}&\text{x}\\\text{y}+\text{x}+\text{x}+\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $\big[\because\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\big]$
$=2\begin{vmatrix}(\text{y}+\text{z})&\text{z}&\text{y}\$\text{z}+\text{x})&\text{z}+\text{x}&\text{x}\$\text{x}+\text{y})&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$\big[\text{Taking 2 common from C}_1\big]$
$=2\begin{vmatrix}\text{y}&\text{z}&\text{y}\\0&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_2]$
$=2\begin{vmatrix}0&\text{z}-\text{x}&-\text{x}\\0&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_3]$
$=2\big[\text{y}(\text{xz}-\text{x}^2+\text{xz}+\text{x}^2)\big]$
$=4\text{xyz}=\text{R.H.S}$
Hence proved.

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