Question

Using the property of determinants and without expanding, prove that

|(y + k, y, y), (y, y + k, y), (y, y, y + k)| = k²(3y + k)

Answer 1

(using C₁ → C₁ + C₂ + C₃)

[take out (5x + 4) common from C₁].

(Using R₂ → R₂ – R₁ and R₃ → R₃ – R₁)

Expanding along C₃, we get 

= (3y + k)[1(k² - 0)]

= k²(3y + k) = RHS