Let A = [(cosθ 0 -sinθ), (0 1 0), (sinθ 0 cosθ)]. If for some θ ∈ (0, π), A^2=A^T,

Question

Let \(A=\left[\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right].\) If for some \(\theta \in(0, \pi), A^{2}=A^{T},\) 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 _____.

Answer 1

Answer is: 6 

\(\because \mathrm{A}\) is orthogonal matrix

\(\therefore A^{T}=A^{-1}\)

\(\Rightarrow \mathrm{A}^{2}=\mathrm{A}^{-1} \quad \left(\because A^{2}=A^{T}\right)\)

\(\Rightarrow A^{3}=I\)

let \(\mathrm{B}=(\mathrm{A}+\mathrm{I})^{3}+(\mathrm{A}-\mathrm{I})^{3}-6 \mathrm{~A}\)

\(=2\left(\mathrm{~A}^{3}+3 \mathrm{~A}\right)-6 \mathrm{~A}\)

\(=2 \mathrm{~A}^{3}\)

\(B=2 I=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\)

Now sum of diagonal elements = 2 + 2 + 2 = 6