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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS4 Marks
Question
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$ 
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$

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