Matrices and Determinants 1 Question 8
8. If $P=\left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right], \quad A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $Q=P A P^T$, then $P^T Q^{2005} P$ is
(2005, 1M)
(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) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
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Answer:
Correct Answer: 8. (a)
Solution:
- 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] $
$ \begin{aligned} & \Rightarrow \quad P^T P=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & \Rightarrow \quad P^T P=I \\ & \Rightarrow \quad P^T=P^{-1} \\ & \end{aligned} $
Since, $\quad Q=P A P^T$
$ \begin{aligned} & \therefore P^T Q^{2005} P=P^T\left[\left(P A P^T\right)\left(P A P^T\right) \ldots 2005 \text { times }\right] P \\ &=\underbrace{\left(P^T P\right) A\left(P^T P\right) A\left(P^T P\right) \ldots\left(P^T P\right) A\left(P^T P\right)}_{2005 \text { times }} \\ &=I A^{2005}=A^{2005} \\ & \therefore \quad A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \\ & 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] \end{aligned} $
$\begin{aligned} A^{2005} & =\left[\begin{array}{cc} 1 & 2005 \\ 0 & 1 \end{array}\right] \\ P^T Q^{2005} P & =\left[\begin{array}{cc} 1 & 2005 \\ 0 & 1 \end{array}\right] \end{aligned}$
