Question

Prove that :

\[\begin{vmatrix}a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c\end{vmatrix} = 2\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}\]

 

Answer 1

\[\text{ Let }\Delta = \begin{vmatrix} a + b & b + c & c + a \\b + c & c + a & a + b\\c + a & a + b & b + c \end{vmatrix}\] 

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