Let A=[cos θ, 0, -sin θ; 0, 1, 0; sin θ, 0, cos θ]. If for some θ ∈(0, π), A²=Aᵀ, then the sum of the diagonal elements of the matrix (A+I)³+(A-I)³-6 A is equal to _________ .

Solution

Maths sir · 4 steps

Do not hunt for θ. Multiply A² = Aᵀ by A to get A³ = I, and the whole expression collapses.

1. A is orthogonal. The columns of A are unit vectors at right angles to each other, so Aᵀ A = I. Multiply A² = Aᵀ on the right by A: you get A³ = Aᵀ A = I.
2. Expand both cubes. A and I commute, so the usual binomial expansion works. In the sum, the ± 3A² terms and the ± I terms cancel.

   (A+I)³ + (A-I)³ = 2A³ + 6A

3. Subtract 6A. The expression becomes 2A³ + 6A - 6A = 2A³ = 2I. Its diagonal is 2, 2, 2.
4. So. The sum of the diagonal elements is 2 + 2 + 2 = 6.

The common slip: 3. Writing the sum of the cubes as A³ + 6A instead of 2A³ + 6A leaves I, whose trace is 3, not 6.

The idea to keep. When a matrix condition looks like A² = Aᵀ, multiply it by A. For an orthogonal matrix, Aᵀ A = I turns it into a clean power rule like A³ = I.

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