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

Answer: (A)

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

Given,

$Q=PA{P}^T$ and $x=P^T{Q}^{2005}P$

Here we compute $P{P}^T$ and on computation we obtain

$P{P}^T=I$

$x=P^T{Q}^{2005}P$

$x=P^T(PA{P}^T{)}^{2005}P$

$x=P^T\left(PA{P}^T\right)(PA{P}^T{)}^{2004}P$

since $P{P}^T=I$

$x=A{P}^T(PA{P}^T{)}^{2004}P$

$x=A{P}^T\left(PA{P}^T\right)(PA{P}^T{)}^{2003}P$

$x=A^2{P}^T(PA{T}^T{)}^{2003}P$

Keep repeating this process, finally we'll be left with $A^{2005}$

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

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

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