Question

If a, b, c, are non zero complex numbers satisfying a² + b² + c² = 0 and `|(b^2 + c^2, ab, ac),(ab, c^2 + a^2, bc),(ac, bc, a^2 + b^2)|` = ka²b²c², then k is equal to ______.

Options

• 1

• 3

• 4

• 2

Answer 1

If a, b, c, are non zero complex numbers satisfying a² + b² + c² = 0 and `|(b^2 + c^2, ab, ac),(ab, c^2 + a^2, bc),(ac, bc, a^2 + b^2)|` = ka²b²c², then k is equal to 4.

Explanation:

Let Δ = `|(b^2 + c^2, ab, ac),(ab, c^2 + a^2, bc),(ac, bc, a^2 + b^2)|`

Multiply C₁ by a, C₂ by b and C₃ by c and hence divide by abc.

= `1/(abc)|(a(b^2 + c^2), ab^2, ac^2),(a^2b, b(c^2 + a^2), bc^2),(a^2c, b^2c, c(a^2 + b^2))|`

Take out a, b, c common from R₁, R₂ and R₃ respectively.

∴ Δ = `(abc)/(abc)|(b^2 + c^2, b^2, c^2),(a^2, c^2 + a^2, c^2),(a^2, b^2, a^2 + b^2)|`

Applying C₁ `rightarrow` C₁ – C₂ – C₃

Δ = `|(0, b^2, c^2),(-2c^2, c^2 + a^2, c^2),(-2b^2, b^2, a^2 + b^2)|`

=  `-2|(0, b^2, c^2),(c^2, c^2 + a^2, c^2),(b^2, b^2, a^2 + b^2)|`

Applying C₂ `rightarrow` C₂ – C₁ and C₃ `rightarrow` C₃ – C₁

= `-2|(0, b^2, c^2),(c^2, a^2, 0),(b^2, 0, a^2)|`

= –2[–b²(c²a²) + c²(–a²b²)]

= 2a²b²c² + 2a²b²c²

= 4a²b²c² 

But Δ = ka²b²c²

∴ k = 4