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If $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] a n d Q = P A P^{T}, t h e n P^{T} (Q^{2005}) P$ is equal to

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1200501

3/2200510

120053/21

13/202005

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detailed solution

Correct option is A

We have,Q = P A PT⇒ PT Q = APT (∵as PTP = I = PPT)∴ PT Q2005 P = A PT Q2004 P= A2 PT Q2003 P = A3 PT Q2002 P= A2004 PT (QP)= A2004 PT (PA)(Q = P A PT ⇒ Q P = P A) = A2005= 1200501

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If P=3212-1232, A=1101 and Q=PAPT, then PT(Q2005)P is equal to

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If $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] a n d Q = P A P^{T}, t h e n P^{T} (Q^{2005}) P$ is equal to