Question

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

• 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

The correct option is A $\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

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