Let P=left[begin{array}{cc}frac{sqrt{3}}{2} & | vedclass

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(B) Given $Q = PAP^{T}$.We need to evaluate $P^{T}Q^{2007}P$.Since $Q = PAP^{T}$,then $Q^{2007} = (PAP^{T})(PAP^{T})\dots(PAP^{T})$ ($2007$ times).$Q^

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$2005$

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(B) Given $Q = PAP^{T}$.We need to evaluate $P^{T}Q^{2007}P$.Since $Q = PAP^{T}$,then $Q^{2007} = (PAP^{T})(PAP^{T})\dots(PAP^{T})$ ($2007$ times).$Q^{2007} = PA(P^{T}P)A(P^{T}P)A\dots A P^{T}$.Since $P$ is an orthogonal matrix,$P^{T}P = I$.Thus,$Q^{2007} = PA^{2007}P^{T}$.Substituting this into the expression,we get $P^{T}(PA^{2007}P^{T})P = (P^{T}P)A^{2007}(P^{T}P) = I \cdot A^{2007} \cdot I = A^{2007}$.For $A = \left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$,we have $A^{n} = \left[\begin{array}{ll}1 & n \ 0 & 1\end{array}ight]$.Therefore,$A^{2007} = \left[\begin{array}{cc}1 & 2007 \ 0 & 1\end{array}ight] = \left[\begin{array}{ll} a & b \ c & d \end{array}ight]$.This gives $a=1, b=2007, c=0, d=1$.Calculating $2a+b-3c-4d = 2(1) + 2007 - 3(0) - 4(1) = 2 + 2007 - 4 = 2005$.

Column $I$	Column $II$
$(A)$ $a+b+c \neq 0$ and $a^2+b^2+c^2=ab+bc+ca$	$(p)$ The equations represent planes meeting only at a single point.
$(B)$ $a+b+c=0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(q)$ The equations represent the line $x=y=z$.
$(C)$ $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq ab+bc+ca$	$(r)$ The equations represent identical planes.
$(D)$ $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$	$(s)$ The equations represent the whole of the three-dimensional space.

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