Let α be a solution of x² + x + 1 = 0, and for some a and b in
R, [4, a, b] [1, 16, 13; -1, -1, 2; -2, -14, -8] = [0, 0, 0]. If 4/α⁴ + m/αᵃ + n/αᵇ = 3, then m + n is equal to _______

Solution

Maths sir · 4 steps

Find a and b from the matrix first, then cut every power of α down using α³ = 1.

1. Find a and b. Multiply the row by the first two columns: 4 - a - 2b = 0 and 64 - a - 14b = 0. Subtracting gives 60 = 12b, so b = 5 and a = -6. The third column checks out: 52 - 12 - 40 = 0.
2. Reduce the powers. α is a non-real cube root of unity, so α³ = 1. That gives 1/α⁴ = 1/α = α², 1/α⁻⁶ = α⁶ = 1 and 1/α⁵ = 1/α² = α.

   4α² + m + nα = 3

3. Remove alpha squared. Put α² = -1 - α: -4 - 4α + m + nα = 3, so (m - 7) + (n - 4)α = 0. Since α is not real and m, n are real, both brackets must be zero.
4. So. m = 7 and n = 4, giving m + n = 11, option (a).

The idea to keep. With ω, first bring every power down using ω³ = 1, then replace ω² by -1-ω. Now p + qω = 0 with real p, q forces p = q = 0.

Answer from the NTA final answer key. Solution by PrepWiser.