Let $A=\left[\begin{array}{ccc}\cos heta & 0 & -\sin heta \ 0 & 1 & 0 \ \sin heta & 0 & \cos heta\end{array} ight]$. If for some $ heta \in(0, \pi)$, $A^2=A^T$, then the sum of the diagonal elements of the matrix $(\mathrm{A}+\mathrm{I})^3+(\mathrm{A}-\mathrm{I})^3-6 \mathrm{~A}$ is…

Question

Let $A=\left[\begin{array}{ccc}\cos 	heta & 0 & -\sin 	heta \ 0 & 1 & 0 \ \sin 	heta & 0 & \cos 	heta\end{array}ight]$. If for some $	heta \in(0, \pi)$, $A^2=A^T$, then the sum of the diagonal elements of the matrix $(\mathrm{A}+\mathrm{I})^3+(\mathrm{A}-\mathrm{I})^3-6 \mathrm{~A}$ is equal to _____ .

Aicharya · Accepted Answer

$\because \mathrm{A}$ is orthogonal matrix
$\begin{aligned}
& \therefore A^T=A^{-1} \\ & \Rightarrow A^2=A^{-1} \left(\because \mathrm{A}^2=\mathrm{A}^{\mathrm{T}}\right) \\ & \Rightarrow A^3=I \\ & \text { let } B=(A+I)^3+(A-I)^3-6 A \\ & =2\left(A^3+3 A\right)-6 A \\ & =2 A^3 \\ & B=2 I=\left[\begin{array}{lll}
2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2
\end{array}\right]
\end{aligned}$
Now sum of diagonal elements $=2+2+2=6$

Let $A=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos…

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