Question

$IfA=\left[\begin{array}{cc}cos\theta & sin\theta \\ sin\theta & -cos\theta \end{array}\right],B=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right],C=AB{A}^T,then{A}^T{C}^{n}Aequalsto\left(nϵZ^{+}\right)$

• A. $\left[\begin{array}{cc}-n& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}1& -n\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}0& 1\\ 1& -n\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 0\\ -n& 1\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}1& 0\\ -n& 1\end{array}\right]$

$A=\left[\begin{array}{cc}cos\theta & sin\theta \\ sin\theta & -cos\theta \end{array}\right]A{A}^T=I\left(i\right)Now,C=AB{A}^T\Rightarrow A^{T}C=B{A}^T\left(ii\right)Now{A}^T{C}^{n}A=A^{T}C.C^{n-1}A=B{A}^T{C}^{n-1}A\left(from\right(ii\left)\right)=B{A}^{T}C.C^{n-2}A=B^2{A}^T{C}^{n-2}A=.......=B^{n-1}{A}^{T}CA=B^{n-1}B{A}^{T}A=B^n=\left[\begin{array}{cc}1& 0\\ -n& 1\end{array}\right]$