Question

if $p=\left[\begin{array}{cc}\sqrt{3}/2& 1/2\\ -1/2& \sqrt{3}/2\end{array}\right],A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T,then$
$P^T{Q}^{2005}P$ is

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

Answer 1

Answer: (A)

Now, $P^{T}P=\left[\begin{array}{cc}\sqrt{3}/2& -1/2\\ 1/2& \sqrt{3}/2\end{array}\right]\left[\begin{array}{cc}\sqrt{3}/2& 1/2\\ -1/2& \sqrt{3}/2\end{array}\right]$
$\Rightarrow P^{T}P=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\Rightarrow P^{T}P=I\Rightarrow P^T=P^{-1}$
Since $Q=PA{P}^T$
$\therefore P^T{Q}^{2005}P=P^T[(PA{P}^T)(PA{P}^T).....2005\text{ times}]P$
$=\underset{2005\text{ times }}{\underbrace{\left(P^{T}P\right)A\left(P^{T}P\right)A\left(P^{T}P\right)\dots \left(P^{T}P\right)A\left(P^{T}P\right)}}$
$=I{A}^{2005}=A^{2005}$
$\therefore A^1\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
$A^2\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
$A^3=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$
.... ..... ..... ........ . ........
.... ...... ..... ....... .......
$A^{2005}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$
$\therefore P^T{Q}^{2005}P=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$