If $$P=\left[ \begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \ \end{matrix} \right],\,A=\left[ \begin{matrix} 1 & 1 \ 0 & 1 \ \end{matrix} \right]$$ and $$Q=PA{{P}^{T}}$$, then $$P({{Q}^{2005}}){{P}^{T}}$$ equal to [IIT Screening 2005]

Question

If $$P=\left[ \begin{matrix}    \frac{\sqrt{3}}{2} & \frac{1}{2}  \    -\frac{1}{2} & \frac{\sqrt{3}}{2}  \ \end{matrix} \right],\,A=\left[ \begin{matrix}    1 & 1  \    0 & 1  \ \end{matrix} \right]$$ and $$Q=PA{{P}^{T}}$$, then $$P({{Q}^{2005}}){{P}^{T}}$$ equal to [IIT Screening 2005]

TuteeHUB · Accepted Answer

$$\left[ \begin{matrix}    \sqrt{3}/2 & 2005  \    1 & 0  \ \end{matrix} \right]$$

TuteeHUB · Answer

$$\left[ \begin{matrix}    1 & 2005  \    0 & 1  \ \end{matrix} \right]$$

TuteeHUB · Answer

$$\left[ \begin{matrix}    1 & 2005  \    \sqrt{3}/2 & 1  \ \end{matrix} \right]$$

TuteeHUB · Answer

$$\left[ \begin{matrix}    1 & \sqrt{3}/2  \    0 & 2005  \ \end{matrix} \right]$$

1.	If \[P=\left[ \begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \end{matrix} \right],\,A=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\] and \[Q=PA{{P}^{T}}\], then \[P({{Q}^{2005}}){{P}^{T}}\] equal to [IIT Screening 2005]
A.	\[\left[ \begin{matrix} 1 & 2005 \\ 0 & 1 \\ \end{matrix} \right]\]
B.	\[\left[ \begin{matrix} \sqrt{3}/2 & 2005 \\ 1 & 0 \\ \end{matrix} \right]\]
C.	\[\left[ \begin{matrix} 1 & 2005 \\ \sqrt{3}/2 & 1 \\ \end{matrix} \right]\]
D.	\[\left[ \begin{matrix} 1 & \sqrt{3}/2 \\ 0 & 2005 \\ \end{matrix} \right]\]
Answer» B. \[\left[ \begin{matrix} \sqrt{3}/2 & 2005 \\ 1 & 0 \\ \end{matrix} \right]\]

If \[P=\left[ \begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \end{matrix} \right],\,A=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\] and \[Q=PA{{P}^{T}}\], then \[P({{Q}^{2005}}){{P}^{T}}\] equal to [IIT Screening 2005]

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