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Question
Prove that :
Solution
\[\text{ Let }\Delta = \begin{vmatrix} a + b & b + c & c + a \\b + c & c + a & a + b\\c + a & a + b & b + c \end{vmatrix}\]
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
\[\Delta = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} + \begin{vmatrix} b & c & a \\c & a & b\\a & b & c \end{vmatrix} \]
\[ = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} + \left( - 1 \right)\begin{vmatrix} a & c & b \\b & a & c\\c & b & a \end{vmatrix}\left[\text{ Applying }C_1 \leftrightarrow C_3\text{ in second determinant to get negative value of the deteminant }\right]\]
\[ = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} | + \left( - 1 \right)\left( - 1 \right) \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} \left[\text{ Applying }C_2 \leftrightarrow C_3 \right]\]
\[ = 2 \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} = RHS\]
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