Let A=pmatrix&0&-0&1&0&0&pmatrix, (0,), and A^2=A^T.

Question

Let $A=\begin{pmatrix}\cos	heta&0&-\sin	heta\0&1&0\\sin	heta&0&\cos	heta\end{pmatrix}$, $	heta\in(0,\pi)$, and $A^2=A^T$. Then the sum of the diagonal elements of $(A+I)^3+(A-I)^3-6A$ is:

Answer: 6

Accepted Answer

$A$ is orthogonal: $AA^T=A^TA=I$, so $A^T=A^{-1}$. Given $A^2=A^T=A^{-1}$, multiplying both sides by $A$ gives $A^3=I$.

Using the identity $(A+I)^3+(A-I)^3=2A^3+6A$:
$$\operatorname{tr}\big[(A+I)^3+(A-I)^3-6A\big]=\operatorname{tr}(2A^3+6A-6A)=\operatorname{tr}(2A^3)=\operatorname{tr}(2I)=6.$$

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