Q111 / 170 · Manipal (MET) 2016 · Mathematics
If $P=\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $Q=P A P^{\prime}$, then $P^{\prime} Q^{2005} P$ is
- A$\left[\begin{array}{cc}1 & 1 \\ 2005 & 1\end{array}\right]$
- Correct answerB$\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
- C$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- D$\left[\begin{array}{cc}1 & 2005 \\ 2005 & 1\end{array}\right]$
Solution
We have,
$P^{\prime} P=\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right]\left[\begin{array}{cc}\sqrt{3} / 2 & -1 / 2 \\ 1 / 2 & \sqrt{3} / 2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]$
$\begin{array}{lrl}\Rightarrow & P^{\prime} P & =I \text { or } P^{\prime}=P^{-1} \\ \text { As, } & Q & =P A P^{\prime}\end{array}$
$\therefore P^{\prime} Q^{2005} P=P^{\prime}\left[\left(P A P^{\prime}\right)\left(P A P^{\prime}\right) \ldots 2005\right.$ times $] P$
$=\underbrace{\left(P^{\prime} P\right) A\left(P^{\prime} P\right) A\left(P^{\prime} P\right) \ldots\left(P^{\prime} P\right) A\left(P^{\prime} P\right)}_{2005 \text { times }}$
$=I A^{2005}=A^{2005}$
Now, $\quad A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]^{\prime} A^2=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$
$A^3=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \ldots A^{2005}$
$=\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
$\therefore P^{\prime} Q^{2005} P=\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
Hence, the correct answer is option: 'B'.