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

Answer 1

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

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

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

$P{P}^T=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$P{P}^1=I$

$\therefore P^1=P^{-1}$

Now,

$Q=PA{P}^T$

$\to P^{T}QP=P^T\left(PA{P}^T\right)P$

$\to P^{T}QP=\left(P^{T}P\right)A\left(P^{T}P\right)$

$\Rightarrow P^{T}QP=IAI=A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

And,

$Q^2=\left(PA{P}^T\right)\left(PA{P}^T\right)$

$=PA\left(P^{T}P\right)\left(A{P}^T\right)$

$=P{A}^2{P}^T$

$\therefore P^T{Q}^{2}P=A^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

$\therefore P^T{Q}^{2015}P=A^{2015}=\left[\begin{array}{cc}1& 2015\\ 0& 1\end{array}\right]$