Let α be a root of the equation x² + x + 1 = 0 and the
matrix A = 1/√3[ 1, 1, 1; 1, α, α ²; 1, α ², α ⁴ ]
then the matrix A³¹ is equal to

Solution

Maths sir · 4 steps

Square A once: A² turns out to be a swap matrix, so A⁴ = I and the powers repeat every 4.

1. Use cube roots. A root of x² + x + 1 = 0 is a complex cube root of unity. So α³ = 1, α⁴ = α and 1 + α + α² = 0.
2. Square the matrix. Each entry of A² is a row of the bracket matrix times a column, divided by 3. Row 1 with column 1 gives 1 + 1 + 1 = 3. Row 2 with column 2 gives 1 + α² + α⁴ = 1 + α² + α = 0. Row 2 with column 3 gives 1 + α³ + α⁶ = 3.

   A² = 1/3(3, 0, 0; 0, 0, 3; 0, 3, 0) = (1, 0, 0; 0, 0, 1; 0, 1, 0)

3. Square again. A² only swaps the second and third rows. Swapping twice puts everything back, so A⁴ = (A²)² = I₃.
4. So. Since 31 = 4 × 7 + 3, we get A³¹ = (A⁴)⁷ A³ = A³, which is option (d).

The idea to keep. For a high power of a matrix, compute the first few powers until you hit I. Once Aᵏ = I, divide the exponent by k and keep only the remainder.

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