Question

If $P=\left\{\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right\}A=\left\{\begin{array}{cc}1& 1\\ 0& 1\end{array}\right\}$ and $Q=PA{P}^T$ and $x=P^T{Q}^{2005}P$, then $x$ is equal to

• A. $\left\{\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right\}$
• B. $\left\{\begin{array}{cc}4+2005\sqrt{3}& 6015\\ 2005& 4-2005\sqrt{3}\end{array}\right\}$
• C. $\frac{1}{4}\left[\begin{array}{cc}2+\sqrt{3}& 1\\ -1& 2-\sqrt{3}\end{array}\right]$
• D. $\frac{1}{4}\left\{\begin{array}{cc}2005& 2-\sqrt{3}\\ 2+\sqrt{3}& 2005\end{array}\right\}$

Answer 1

The correct option is A $\left\{\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right\}$

$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

$P^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

Now, $P^{T}P=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

$p^T{Q}^{2005}P=P^T(PA{P}^T{)}^{2005}P=P^T(PA{P}^T\left)\right(PA{P}^T\left)\right(PA{P}^T)....P=A^{2005}$

Here,

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

$A^2=A.A$

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

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

$A^3=A^2.A$

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

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

$\therefore$ $A^{2005}=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

$\therefore$ $P^T{Q}^{2005}P=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$

$X=\left[\begin{array}{cc}1& 2005\\ 0& 1\end{array}\right]$