Let P = [ cases 3 2 & (1)/(2) - (1)/(2) & 3 2 cases ], A = [ cases...

Given:

P = \left[\begin{cases}\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ - \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{cases}\right]

∴ P{P}^{T} = \left[\begin{cases}\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ - \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{cases}\right]\left[\begin{cases}\dfrac{\sqrt{3}}{2} & - \dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{cases}\right] = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right]

$⇒ P{P}^{T} = I$

Now,

${P}^{T}{\left(PA{P}^{T}\right)}^{2007}P = {P}^{T}\mathop{\mathop{\left(PA{P}^{T}\right)\left(PA{P}^{T}\right)\left(PA{P}^{T}\right) . . . \left(PA{P}^{T}\right)}_{⏟}}_{2007times}P$

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

Now,

A = \left[\begin{cases}1 & 1 \\ 0 & 1\end{cases}\right]{A}^{2} = \left[\begin{cases}1 & 2 \\ 0 & 1\end{cases}\right]{A}^{3} = \left[\begin{cases}1 & 3 \\ 0 & 1\end{cases}\right] ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ {A}^{2007} = \left[\begin{cases}1 & 2007 \\ 0 & 1\end{cases}\right]

So,

⇒ {P}^{T}{\left(PA{P}^{T}\right)}^{2007}P = \left[\begin{cases}1 & 2007 \\ 0 & 1\end{cases}\right]

⇒ {P}^{T}QP = \left[\begin{cases}1 & 2007 \\ 0 & 1\end{cases}\right] = \left[\begin{cases}a & b \\ c & d\end{cases}\right]

So,

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

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