Question
If \[P = \left[ \begin{array}{l}
\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\frac{1}{2}\\
- \frac{1}{2}\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]{\rm{and}}\] \[A = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,1\\
0\,\,\,\,\,\,1
\end{array} \right]{\rm{ and}}\] $$Q = PA{P^T}{\text{ and }}x = {P^T}{Q^{2005}}P$$ then $$x$$ is equal to
A.
\[\left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,\,2005\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,1
\end{array} \right]\]
B.
\[\left[ \begin{array}{l}
4 + 2005\sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,6015\\
\,\,\,\,\,2005\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 - 2005\sqrt 3
\end{array} \right]\]
C.
\[\frac{1}{4}\left[ \begin{array}{l}
2 + \sqrt 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 - \sqrt 3
\end{array} \right]\]
D.
\[\frac{1}{4}\left[ \begin{array}{l}
2005\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 - \sqrt 3 \\
2 + \sqrt 3 \,\,\,\,\,\,\,\,\,2005
\end{array} \right]\]
Answer :
\[\left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,\,2005\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,1
\end{array} \right]\]
Solution :
\[\begin{array}{l}
{\rm{Now,\, }}{P^T}P = \left[ \begin{array}{l}
\frac{{\sqrt 3 }}{2}\,\,\,\,\, - \frac{1}{2}\\
\frac{1}{2}\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\left[ \begin{array}{l}
- \frac{{\sqrt 3 }}{2}\,\,\,\,\,\frac{1}{2}\\
- \frac{1}{2}\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\\
\Rightarrow {P^T}P = \left[ \begin{array}{l}
1\,\,\,\,\,1\\
0\,\,\,\,\,1
\end{array} \right]\\
\Rightarrow {P^T}P = I\\
\Rightarrow {P^T} = {P^{ - I}}\\
{\rm{Since,\, }}Q = PA{P^T}\\
\therefore \,{P^T}{Q^{2005}}P = {P^T}\left[ {\left( {PA{P^T}} \right)\left( {PA{P^T}} \right)......\,2005{\rm{ \,times}}} \right]P\\
= \frac{{\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)......\left( {PA{P^T}} \right)A\left( {PA{P^T}} \right)}}{{2005{\rm{ \,times}}}}\\
= I{A^{2005}} = {A^{2005}}\\
\therefore \,{A^1} = \left[ \begin{array}{l}
1\,\,\,\,\,1\\
0\,\,\,\,\,1
\end{array} \right]\\
{A^2} = \left[ \begin{array}{l}
1\,\,\,\,\,1\\
0\,\,\,\,\,1
\end{array} \right]\left[ \begin{array}{l}
1\,\,\,\,\,1\\
0\,\,\,\,\,1
\end{array} \right] = \left[ \begin{array}{l}
1\,\,\,\,\,2\\
0\,\,\,\,\,1
\end{array} \right]\\
{A^3} = \left[ \begin{array}{l}
1\,\,\,\,\,2\\
0\,\,\,\,\,1
\end{array} \right]\left[ \begin{array}{l}
1\,\,\,\,\,1\\
0\,\,\,\,\,1
\end{array} \right] = \left[ \begin{array}{l}
1\,\,\,\,\,3\\
0\,\,\,\,\,1
\end{array} \right]\\
{A^{2005}} = \left[ \begin{array}{l}
1\,\,\,\,\,2005\\
0\,\,\,\,\,\,\,\,1
\end{array} \right]\\
\therefore \,{P^T}{Q^{2005}}P = \left[ \begin{array}{l}
1\,\,\,\,\,2005\\
0\,\,\,\,\,\,\,\,1
\end{array} \right]\\
{\rm{Therefore\, option\, A\, is\, the\, right\, answer}}{\rm{.}}
\end{array}\]