Let P=[ … ] and Q=PAP^T. If P^T Q^2007P=[ … ]then 2a+b-3c-4d is equal to

Given:

$$P=[\begin{matrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{matrix}]$$P=[23​​−21​​21​23​​​]

$$\therefore P{P}^{T}=[\begin{matrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{matrix}][\begin{matrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{matrix}]=[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}]$$∴PPT=[23​​−21​​21​23​​​][23​​21​​−21​23​​​]=[10​01​]

$$\Rightarrow P{P}^{T}=I$$⇒PPT=I

Now,

$${P}^{T}{(PA{P}^{T})}^{2007}P={P}^{T}\underset{2007\mathrm{times}}{\underset{⏟}{(PA{P}^{T})(PA{P}^{T})(PA{P}^{T})...(PA{P}^{T})}}P$$PT(PAPT)2007P=PT2007times⏟(PAPT)(PAPT)(PAPT)...(PAPT)​​P

$$\Rightarrow {P}^{T}{(PA{P}^{T})}^{2007}P={A}^{2007}$$⇒PT(PAPT)2007P=A2007

Now,

$$A=[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}] {A}^{2}=[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}] {A}^{3}=[\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}] \vdots \vdots \vdots \vdots \vdots \vdots \vdots \vdots {A}^{2007}=[\begin{matrix}1 & 2007 \\ 0 & 1\end{matrix}]$$A=[10​11​]A2=[10​21​]A3=[10​31​]⋮⋮⋮⋮⋮⋮⋮⋮A2007=[10​20071​]

So,

$$\Rightarrow {P}^{T}{(PA{P}^{T})}^{2007}P=[\begin{matrix}1 & 2007 \\ 0 & 1\end{matrix}]$$⇒PT(PAPT)2007P=[10​20071​]

$$\Rightarrow {P}^{T}QP=[\begin{matrix}1 & 2007 \\ 0 & 1\end{matrix}]=[\begin{matrix}a & b \\ c & d\end{matrix}]$$⇒PTQP=[10​20071​]=[ac​bd​]

So,

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

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