If a ≠ b ≠ c and |(a,b,c)(b,c,a)(c,a,b)|=0 then using properties of determinants, prove that a + b + c = 0.

Question

If a ≠ b ≠ c \(\begin{vmatrix} a & b & c \\[0.3em] b & c & a \\[0.3em] c & a & b \end{vmatrix}=0\) and then using properties of determinants, prove that a + b + c = 0.

Answer 1

We have,

\(\begin{vmatrix} a & b & c \\[0.3em] b & c & a \\[0.3em] c & a & b \end{vmatrix}=0\)

Applying C₁ \(\longrightarrow\)C₁+ C₂ +C₃,we get

Taking (a + b + c) common from C₁, we get

Applying R₂ \(\longrightarrow\) R₂ - R₁ and R₃ \(\longrightarrow\) R₃ - R₁, we get 

Expanding along C₁, we get