Question

If $P=\left[\begin{array}{cc}\cos (\pi /6)& \sin (\pi /6)\\ -\sin (\pi /6)& \cos (\pi /6)\end{array}\right],A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PAP\varphi$ then $P\varphi Q^{2007}P$ is equal to

• A. $\left[\begin{array}{cc}1& 2007\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& \sqrt{3}/2\\ 0& 2007\end{array}\right]$
• C. $\left[\begin{array}{cc}\sqrt{3}/2& 2007\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}\sqrt{3}/2& -1/2\\ 1& 2007\end{array}\right]$

Answer 1

Answer: (A)

Note that $P^{\prime }=P^{-1}$ Now, $Q=PA{P}^{\prime }=PA{P}^{-1}$
$\Rightarrow Q^{2007}=P{A}^{2007}{P}^{-1}$
$\therefore P^{\prime }{Q}^{2007}P=P^{-1}\left(P{A}^{2007}{P}^{-1}\right)P=A^{2007}=(I+B{)}^{2007}$
where $B=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$.
As $B^2=0$, we get $B^r=0\forall r\ge 2$.
Thus, by binomial theorem,
$A^{2007}=I+2007B=\left[\begin{array}{cc}1& 2007\\ 0& 1\end{array}\right]$