Let $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T$. If $P^T{Q}^{2007} P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$then $2a+b-3c-4d$ is equal to

Given:

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

$\therefore P{P}^T=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& -\frac{1}{2}\\ \frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow P{P}^T=I$

Now,

$P^T{\left(PA{P}^T\right)}^{2007}P=P^T\underset{2007 \mathrm{times}}{\underbrace{\left(PA{P}^T\right)\left(PA{P}^T\right)\left(PA{P}^T\right)...\left(PA{P}^T\right)}}P$

$\Rightarrow P^T{\left(PA{P}^T\right)}^{2007}P=A^{2007}$

Now,

$$\begin{gathered} A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right] \\ {A}^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right] \\ {A}^3=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right] \\ ⋮ ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ ⋮ \\ {A}^{2007}=\left[\begin{array}{cc}1& 2007\\ 0& 1\end{array}\right] \end{gathered}$$

So,

$\Rightarrow P^T{\left(PA{P}^T\right)}^{2007}P=\left[\begin{array}{cc}1& 2007\\ 0& 1\end{array}\right]$

$\Rightarrow P^{T}QP=\left[\begin{array}{cc}1& 2007\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

So,

$a=1, b=2007, c=0, d=1$

$2a+b+3c-4d=2+2007-4=2005$