Question

Let $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $P=\left[\begin{array}{cc}\cos \frac{\pi }{6}& \sin \frac{\pi }{6}\\ -\sin \frac{\pi }{6}& \cos \frac{\pi }{6}\end{array}\right]$ and $Q=P{AP}^T,$ then $P^T{Q}^{2021}P$ is equal to

• A. $\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}0& 2021\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}2021& 0\\ 0& 2021\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 2021\\ 2021& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$

Let, $P^T\cdot P=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\therefore P^{T}P=I$
Since $Q={PAP}^T$
$\Rightarrow P^T{Q}^{2021}P=P^T{\left(P{AP}^T\right)}^{2021}P$ $=P^T\cdot [\left({PAP}^T\right)\left({PAP}^T\right)\cdots 2021\text{times}]\cdot P={IA}^{2021}=A^{2021}$
Now, $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]A^3=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$
Similarly, $A^{2021}=\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$
$\therefore P^T{Q}^{2021}P=\left[\begin{array}{cc}1& 2021\\ 0& 1\end{array}\right]$