Question

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

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

Answer 1

The correct option is D $\left[\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right]$

Given,

$P=\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)$

$P^{T}p=\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)\left(\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right)$

$p^{T}p=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

$⟹P^{T}P=I⟹P^T=P^{-1}$

Also

$Q=PA{P}^T$

$\therefore P^T{Q}^{2015}P=P^T\left[\right(PA{P}^T\left)\right(PA{P}^T)....2015$ timess$]P$

$=I{A}^{2015}$

As $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$

$A^2=A.A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$

$A^3=A^2.A=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$

..........

.........

$A^{2015}=\left(\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right)$

Hence $P^T{Q}^{2015}P=\left(\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right)$