Question

Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to

Hint:

Use the properties of orthogonal matrices to simplify the problem.

Answer 1

Correct Answer: 6

1. Given $A$ is an orthogonal matrix, then $A^T = A^{-1}$. If $A^2 = A^{-1}$, then $A^2 = A^T$.
2. Given $A^2 = A^T$, it follows that $A^3 = I$.
3. To calculate $(A + I)^3 + (A - I)^3 - 6A$, we expand the expression: $(A + I)^3 + (A - I)^3 - 6A = 2(A^3 + 3A) - 6A = 2A^3$. Since $A^3 = I$, the expression simplifies to $2I$.
4. The sum of the diagonal elements of $2I$ is calculated as follows: \[ 2I = \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] The sum of the diagonal elements is $2 + 2 + 2 = 6$. Therefore, the correct answer is (1) 6.