Let A = [ 1/√10, 3/√10; (- 3)/√10, 1/√10 ] and B = [ 1, - i; 0, 1 ], where i = √(- 1). If M=Aᵀ B A, then the inverse of the matrix AM²⁰²³Aᵀ is

Solution

Maths sir · 4 steps

A is orthogonal, so every AAᵀ in the product collapses to I and you are left with B²⁰²³.

1. A is orthogonal. The rows of A are 1/√10(1, 3) and 1/√10(-3, 1). Each has length 1 and their dot product is (-3 + 3)/10 = 0, so AAᵀ = AᵀA = I.
2. Powers of M. M² = AᵀBA AᵀBA = AᵀB(AAᵀ)BA = AᵀB²A. Each middle AAᵀ becomes I in the same way, so M²⁰²³ = AᵀB²⁰²³A, and then AM²⁰²³Aᵀ = (AAᵀ)B²⁰²³(AAᵀ) = B²⁰²³.
3. Powers of B. B² = (1, -2i; 0, 1), and each extra factor of B adds another -i to the top-right entry. So Bⁿ = (1, -ni; 0, 1).

   (B²⁰²³)⁻¹ = (1, -2023i; 0, 1)⁻¹ = (1, 2023i; 0, 1)

4. So. The determinant is 1, so the inverse just flips the sign of the off-diagonal entry. The answer is (1, 2023i; 0, 1), option (c).

The idea to keep. When you see AᵀBA or ABAᵀ, first check whether A is orthogonal. If it is, every AAᵀ in a power collapses to I, and only the middle matrix gets raised to the power.

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