If P = 3 2 1 2 - 1 2 3 2 , A = 1 1 0 1 and Q = P A P T , then P T Q 2015 P is : | TestHub

If $P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}], A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ and $Q = P A P^{T},$ then $P^{T} Q^{2015} P$ is :

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$P = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ - \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}] P^{T} = [\begin{matrix} \frac{\sqrt{3}}{2} & \frac{- 1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{matrix}]$

$P P^{T} = P^{T} P = I$

$Q^{2015} = (P A P^{T}) (P A P^{T})$ ......( $2015$ terms)

$= P A^{2015} P^{T}$ ... $(i)$

$∴ P^{T} Q^{2015} P = P^{T} P A^{2015} P^{T} P = A^{2015}$ from.. $(i)$

Given $A = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$
$\Rightarrow$ $A^{2} = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$

$\Rightarrow$ $A^{3} = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}]$

$∴$ $A^{2015} = [\begin{matrix} 1 & 2015 \\ 0 & 1 \end{matrix}]$

Mathematics - Matrices Question with Solution | TestHub

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