Question

Using properties of determinants, prove that |(3a, -a + b, -a + c), (-b + a, 3b, -b + c), (-c + a, -c + b, 3c)| = 3(a + b + c)(ab + bc + ca)

Answer 1

(using C₁ → C₁ + C₂ + C₃)

(Taking out (a + b + c) common from C₁)

Now applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

Expanding along C₁, we get 

= (a + b + c)[(2b + a) (2c + a) − (a − b) (a − c)] 

= (a + b + c)[4bc + 2ab + 2ac + a − a + ac + ba − bc] 

= (a + b + c) (3ab + 3bc + 3ac) = 3(a + b + c)(ab + bc + ca) = RHS