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(B) Given $A = \begin{bmatrix} \cos 	heta & 0 & -\sin 	heta \ 0 & 1 & 0 \ \sin 	heta & 0 & \cos 	heta \end{bmatrix}$.Since $A$ is an orthogonal

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$6$

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(B) Given $A = \begin{bmatrix} \cos 	heta & 0 & -\sin 	heta \ 0 & 1 & 0 \ \sin 	heta & 0 & \cos 	heta \end{bmatrix}$.Since $A$ is an orthogonal matrix,$A^T A = I$,which implies $A^T = A^{-1}$.Given $A^2 = A^T$,we have $A^2 = A^{-1}$.Multiplying both sides by $A$,we get $A^3 = I$.Now,consider the expression $B = (A + I)^3 + (A - I)^3 - 6A$.Expanding the cubes:$(A + I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 = A^3 + 3A^2 + 3A + I$.$(A - I)^3 = A^3 - 3A^2I + 3AI^2 - I^3 = A^3 - 3A^2 + 3A - I$.Adding these two expressions:$(A + I)^3 + (A - I)^3 = (A^3 + 3A^2 + 3A + I) + (A^3 - 3A^2 + 3A - I) = 2A^3 + 6A$.Substituting this into $B$:$B = (2A^3 + 6A) - 6A = 2A^3$.Since $A^3 = I$,we have $B = 2I = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{bmatrix}$.The sum of the diagonal elements (trace) of $B$ is $2 + 2 + 2 = 6$.

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 . . . . . . .

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