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 (integer)
6
Solution
<p>Note that $A$ is orthogonal:</p>
<p>$A A^T=A^T A=I \text { and } A^T=A^{-1}$</p>
<p>Given $A^2=A^T$, then:</p>
<p>$$\begin{aligned}
& A^3=I \\
& \operatorname{Tr}(A+I)^3+(A-l)^3-6 A=\operatorname{Tr}\left(2 A^3+6 A-6 A\right) \\
& =\operatorname{Tr}\left(2 A^3\right)=\operatorname{Tr}(2 I) \\
& \left(\text { Using }(A+l)^3+(A-I)^3=2 A^3+6 A \text { and } 2 A^3=2 I\right)= \\
& 6
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
This question is part of PrepWiser's free JEE Main question bank. 274 more solved questions on Matrices and Determinants are available — start with the harder ones if your accuracy is >70%.