If A = ( 1/√5, 2/√5; (- 2)/√5, 1/√5 ), B = ( 1, 0; i, 1 ), i = √(- 1), and Q = A T BA, then the inverse of the matrix A Q²⁰²¹ A T is equal to :

Solution

Maths sir · 4 steps

A is orthogonal, so A and Aᵀ cancel and the whole product collapses to B²⁰²¹.

1. A is orthogonal. The rows of A are 1/√5(1, 2) and 1/√5(-2, 1). Each has length 1 and their dot product is 1/5(-2 + 2) = 0, so AAᵀ = AᵀA = I.
2. Collapse the powers. Q² = AᵀBA AᵀBA = AᵀB²A, because the middle AAᵀ is I. Repeating this gives Q²⁰²¹ = AᵀB²⁰²¹A, so AQ²⁰²¹Aᵀ = (AAᵀ)B²⁰²¹(AAᵀ) = B²⁰²¹.
3. Power of B. Write B = I + N with N = (0, 0; i, 0). Since N² = O, the binomial expansion stops after two terms, giving Bⁿ = I + nN.

   B²⁰²¹ = (1, 0; 2021i, 1), (B²⁰²¹)⁻¹ = (1, 0; -2021i, 1)

4. So. The determinant of B²⁰²¹ is 1, and its inverse just flips the sign of the off-diagonal entry. The inverse of AQ²⁰²¹Aᵀ is option (b).

The idea to keep. When you see AᵀBA with an orthogonal A, the A's cancel inside every power: (AᵀBA)ⁿ = AᵀBⁿA. And for I + N with N² = O, (I+N)ⁿ = I + nN.

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