Question

Let $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $P=\left[\begin{array}{cc}\cos \frac{\pi }{6}& \sin \frac{\pi }{6}\\ -\sin \frac{\pi }{6}& \cos \frac{\pi }{6}\end{array}\right],$ $Q=PA{P}^T$ then $P^T{Q}^{2013}P$ is equal to

• A. $\left[\begin{array}{cc}1& 2013\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}0& 2013\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}2013& 0\\ 0& 2013\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 2013\\ 2013& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 2013\\ 0& 1\end{array}\right]$

Let $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

$P=\left[\begin{array}{cc}\cos \pi /6A& \sin \pi /6\\ -\sin \pi /6& \cos \pi /6\end{array}\right]$

$Q=PA{P}^T$

then $P^T{Q}^{2015}P$ is = ?

Now by using property of transpose.

$P^T.P=1$

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

$\therefore P^{T}P=I$

Hence $P^T=P^{-1}$

Since $Q=PA{P}^T$

$P^T{Q}^{2013}P=P^T(PA{P}^T{)}^{2013}P$

i.e. $P^T{Q}^{2013}P=P^T\left[\right(PA{P}^T\left)\right(PA{P}^T)...2013times]P$

$=P^T.P\left[\right(AP{P}^T\left)\right(AP{P}^T)....2013times]$

$=I{A}^{2013}$

$=A^{2013}$

Now $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

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

$A^3=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

Similarly

$A^{2013}=\left[\begin{array}{cc}1& 2013\\ 0& 1\end{array}\right]$

$\therefore P^T{Q}^{2013}P=\left[\begin{array}{cc}1& 2013\\ 0& 1\end{array}\right]$