Question

Using properties of determinants, prove that |(a, a + b, a + b + c), (2a, 3a + 2b, 4a + 3b + 2c), (3a, 6a + 3b, 10a + 6b + 3c)| = a³

Answer 1

[since C₁ = C₂ in second determinant]

[since C₂ = C₃ in second determinant and C₁ = C₃ in third determinant]

Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁ we get

Expanding along R₁ we get 

= a³ .1(7 – 6) – 0 + 0 

= a³ .