Let X=[1; 1; 1] and A=[-1, 2, 3; 0, 1, 6; 0, 0, -1]. For k ∈ N, if X′ Aᵏ X=33, then k is equal to _______.

Solution

Maths sir · 4 steps

Square A first. A² turns out almost the identity, so every even power follows a clean pattern.

1. Read the product. With X a column of 1s, X'MX is the sum of all nine entries of M. So you need the sum of the entries of Aᵏ.
2. Compute A squared. Multiplying out gives A² = (1, 0, 6; 0, 1, 0; 0, 0, 1). Each further multiplication by A² adds another 6 to the top-right entry and leaves the rest unchanged.

   A²ᵐ = (1, 0, 6m; 0, 1, 0; 0, 0, 1), A²ᵐ⁺¹ = (-1, 2, 3-6m; 0, 1, 6; 0, 0, -1)

3. Check both cases. For even k = 2m the entries add up to 3 + 6m. For odd k = 2m+1 they add up to 10 - 6m, which is at most 10. That can never reach 33.
4. So. k must be even, with 3 + 6m = 33. That gives m = 5, so k = 2m = 10.

The common slip: 5. Solving 3 + 6m = 33 and reporting m gives 5. The power is k = 2m, which is 10.

The idea to keep. For a high power of a matrix, compute A² and maybe A³ and look for a pattern. Treat even and odd powers separately when the diagonal has -1s.

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