Question

If $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$, $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T$ then $P^T{Q}^{2015}P$ is :

• A. $\left[\begin{array}{cc}0& 2015\\ 0& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}2015& 0\\ 0& 2015\end{array}\right]$
• C. $\left[\begin{array}{cc}2015& 0\\ 1& 2015\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right]$

Answer 1

Answer: (D)

$P=\left|\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right|P^T=\left|\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right|$
$P{P}^T=P^{T}P=1$
$Q^{2005}=\left(PA{P}^T\right)\left(PA{P}^T\right)................\left(2005terms\right)$
$=P{A}^{2005}{P}^T$
$P^T{\theta }^{2005}P=A^{2005}$
$A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
$A^3=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$
$\therefore A^{2005}=\left[\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right]$