Let P=[[|array|c|c| √32 & (1)/(2) -(1)/(2) & √32 array]]|,A=[array|c|c| 1 & 1 0 & 1 array]…

Question

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

Accepted Answer

$P=[[|\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \ \frac{-1}{2} & \frac{\sqrt{3}}{2} \end{array}]]|,A=[\begin{gathered} 1 1 \ 0 1 ]$

${PP}^{T}=[[|\begin{array}{cc} \frac{\sqrt{3}}{2} \frac{1}{2} \ \frac{-1}{2} \frac{\sqrt{3}}{2} \end{array}]]|[[|\begin{array}{cc} \frac{\sqrt{3}}{2} \frac{-1}{2} \ \frac{1}{2} \frac{\sqrt{3}}{2} \end{array}]]|$

$=[[|\begin{array}{cc} \frac{3}{4}+\frac{1}{4} -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \ \frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \frac{1}{4}+\frac{3}{4} \end{array}]]|=[ 1 0 \ 0 1 ]=I$

Similarly ${P}^{T}P=1$

$\because $ $Q=PA{P}^{T}$

Now, ${Q}^{2007}=(PA{P}^{T})(PA{P}^{T})\ldots 2007$ times $=P{A}^{2007}{P}^{T}$

$\begin{gathered} A=[1 \quad 1 \ 0 \quad 1] \ \Rightarrow {A}^{2}=[1 \quad 1 \ 0 \quad 1][1 \quad 1 \ 0 \quad 1]=[1+0 \quad 1+1 \ 0+0 \quad 0+1]=[1 \quad 2 \ 0 \quad 1] \end{gathered}$

${A}^{3}=[ 1 3 \ 0 1 ]$

.

${A}^{2007}=[ 1 2007 \ 0 1 ]$

$\begin{gathered} {P}^{T}{Q}^{2007}P={A}^{2007}=[1 \quad 2007 \ 0 \quad 1] \ \Rightarrow [a \quad b \ c \quad d]=[1 \quad 2007 \ 0 \quad 1 \end{gathered}] \end{gathered}$

$\begin{gathered} \Rightarrow a=1,b=2007,c=0,d=1 \ 	herefore 2a+b-3c-4d=2(1)+2007-3(0)-4(1) \ =2+2007-4=2005 \end{gathered}$

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