Question

If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]

Answer 1

\[\text{ Let }\Delta = \begin{vmatrix} b + c c + a a + b\\c + a a + b b + c\\a + b b + c c + a \end{vmatrix}\] 
\[ = \begin{vmatrix} 2( a + b + c ) & 2( a + b + c ) & 2( a + b + c)\\c + a & a + b & b + c\\a + b & b + c & c + a \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 + R_2 + R_3 \right]\] 
\[ = 2( a + b + c )\begin{vmatrix}1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix}\] 
\[ = 2( a + b + c )\begin{vmatrix} 1 & 0 & 0\\c + a & b - c & b - a\\a + b & c - a & c - b \end{vmatrix}\left[\text{ Applying }C_2 \to C_2 - C_1\text{ and }C_3 \to C_3 - C_1 \right]\] 
\[ = 2\left( a + b + c \right)\left\{ 1\begin{vmatrix}b - c & b - a \\ c - a & c - b\end{vmatrix} \right\}\] 
\[ = 2\left( a + b + c \right)\left\{ \left( b - c \right)\left( c - b \right) - \left( b - a \right)\left( c - a \right) \right\}\] 
\[ = - 2\left( a + b + c \right)\left\{ a^2 + b^2 + c^2 - ab - bc - ca \right\}\] 
\[ = - \left( a + b + c \right)\left\{ 2 a^2 + 2 b^2 + 2 c^2 - 2ab - 2bc - 2ca \right\}\] 
\[ = - \left( a + b + c \right)\left\{ \left( a - b \right)^2 + \left( b - c \right)^2 + \left( c - a \right)^2 \right\}\] 
\[\text{ But }\Delta = 0 \left[\text{ Given }\right]\] 
\[ \Rightarrow - \left( a + b + c \right)\left\{ \left( a - b \right)^2 + \left( b - c \right)^2 + \left( c - a \right)^2 \right\} = 0\] 
\[ \Rightarrow\text{ Either }\left( a + b + c \right) = 0 or \left( a - b \right)^2 + \left( b - c \right)^2 + \left( c - a \right)^2 = 0\] 
\[ \Rightarrow \left( a + b + c \right) = 0\text{ or }a = b = c\] 

Hence proved.