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 $P^T{Q}^{2005}P$ is equal to

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

Answer 1

Answer: (A)

It is given that $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ \frac{-1}{2}& \frac{\sqrt{3}}{2}\end{array}\right],A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T$
Multiplying by $P^T$ on both sides, we get
$P^{T}Q=A{P}^T\left\{\because P^{T}P=I=P{P}^T\right\}$
$\therefore P^T{Q}^{2005}P=A{P}^T{Q}^{2004}P$ $\left\{\because P^{T}Q=A{P}^T\right\}$
$=A^2{P}^T{Q}^{2003}P$
$=A^3{P}^T{Q}^{2002}P$
$=A^{2004}{P}^{T}QP=A^{2004}{P}^T\left(PA\right)$
$=A^{2004}A=A^{2005}$
$\Rightarrow P^T{Q}^{2005}P=A^{2005}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$