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 $P\left(Q^{2005}\right)P^T$ equal to

• A. $\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}\sqrt{3}/2& 2005\\ 1& 0\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 2005\\ \sqrt{3}/2& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& \sqrt{3}/2\\ 0& 2005\end{array}\right]$

Answer 1

Answer: (A)

Given $Q=PA{P}^T$
$\Rightarrow P^{T}Q=A{P}^T,\left(\because P{P}^T=I\right)$
$\Rightarrow P^T{Q}^{2005}P=A{P}^T{Q}^{2004}P=A{P}^T{Q}^{2003}PA$
$\left(\because Q=PA{P}^T\Rightarrow QP=PA\right)$
$=A{P}^T{Q}^{2002}P{A}^2=A{P}^{T}P{A}^{2004}$
$=AI{A}^{2004}=A^{2005}={\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}^{{}^{2005}}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$