Let A = [(cosθ 0 -sinθ), (0 1 0), (sinθ 0 cosθ)]. Here A^2 = A^T. Then find trace [(A+I)^3 + (A-I)^3 - 6A].

Question

Let A  \( = \begin{bmatrix} \cos \theta & 0 & -\sin\theta \\[0.3em] 0 & 1 & 0 \\[0.3em] \sin \theta & 0 & \cos \theta \end{bmatrix}.\) Here \(A^2=A^T.\) Then find trace \([(A+I)^3 + (A-I^3)-6A].\)

Answer 1

Answer is: 6 

Here, A is orthogonal matrix 

So, \(A^T = A^{-1}\) 

\(\Rightarrow A^2 = A^T \Rightarrow A^2 = A^{-1} \Rightarrow A^3 = I\)

\(B = (A + I) ^ 3 + (A - I) ^ 3 - 6A\)

\(= 2(A^3 + 3A) - 6A\) 

\(=2A^3\)  

\(= 2I\) 

Tr(B) = 2 + 2 + 2 = 6