Question

Using properties of the determinant, show that:

`|(b-c,c-a,a-b),(c-a,a-b,b-c),(2(a-b), 2(b-c), 2(c-a))|`

Answer 1

Taking out common factors

Δ = 2 

`|(b-c,c-a,a-b),(c-a,a-b,b-c),(a-b, b-c,c-a)|`   ...[Taking 2 common from R3]

R1 → R1+ R2 + R3

Δ = 2 `|(0,0,0),(c-a,a-b,b-c),(a-b, b-c,c-a)|`    ...[Applying Row Transformation]

Δ = 2 × 0 = 0

Hence Proved