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 that $A$ is an orthogonal matrix: \[ A^T = A^{-1} \] \[ A^2 = A^{-1} \] 
2. Given $A^2 = A^T$: \[ A^3 = I \] 
3. Calculate $(A + I)^3 + (A - I)^3 - 6A$: \[ (A + I)^3 + (A - I)^3 - 6A = 2(A^3 + 3A) - 6A = 2A^3 = 2I \] 
4. Sum of the diagonal elements of $2I$: \[ 2I = \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \] \[ \text{Sum of diagonal elements} = 2 + 2 + 2 = 6 \] Therefore, the correct answer is (1) 6.

Answer 2

We are given \(A=\begin{bmatrix}\cos\theta&0&-\sin\theta\\[2pt]0&1&0\\[2pt]\sin\theta&0&\cos\theta\end{bmatrix}\) and for some \(\theta\in(0,\pi)\) it satisfies \(A^{2}=A^{T}\). We must find the sum of diagonal entries (the trace) of \((A+I)^3+(A-I)^3-6A\).

Concept Used:

For any matrix \(A\) that commutes with \(I\) (always true), the binomial expansion yields

\[ (A\pm I)^3=A^3\pm 3A^2+3A\pm I. \]

Also, the given \(A\) is a rotation matrix about the \(y\)-axis, hence orthogonal: \(A^T=A^{-1}\). Using the condition \(A^2=A^T\) gives \(A^3=I\).

Step-by-Step Solution:

Step 1: Use the given condition to simplify powers of \(A\).

\[ A^2=A^T \quad\Rightarrow\quad A^3=A\cdot A^2=A\cdot A^T=I\quad(\text{since }A\text{ is orthogonal}). \]

Step 2: Expand and simplify \((A+I)^3+(A-I)^3-6A\).

\[ (A+I)^3+(A-I)^3 = (A^3+3A^2+3A+I) + (A^3-3A^2+3A-I) = 2A^3+6A. \] \[ \Rightarrow\ (A+I)^3+(A-I)^3-6A = 2A^3. \]

Step 3: Use \(A^3=I\) to compute the trace.

\[ 2A^3=2I \quad\Rightarrow\ \operatorname{tr}\big(2A^3\big)=\operatorname{tr}(2I)=2\,\operatorname{tr}(I)=2\cdot 3=6. \]

Final Computation & Result

The sum of the diagonal elements of \((A+I)^3+(A-I)^3-6A\) is 6.