Question

If $P=\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix},A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q=PAP^{T}$ then $P^{T}Q^{2005}P$ is equal to

• A. $\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}$
• B. $\begin{bmatrix} \frac{\sqrt{3}}{2} & 2005 \\ 1 & 0 \end{bmatrix}$
• C. $\begin{bmatrix} 1 & 2005 \\ \frac{\sqrt{3}}{2} & 1 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & \frac{\sqrt{3}}{2} \\ 0 & 2005 \end{bmatrix}$

Answer 1

Answer: (A)

It is given that $P=\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ \frac{- 1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix},A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q=PAP^{T}$
Multiplying by $P^{T}$ on both sides, we get
$P^{T}Q=AP^{T}\left\{\because P^{T} P = I = P P^{T}\right\}$
$\therefore P^{T}Q^{2005}P=AP^{T}Q^{2004}P$ $\left\{\because P^{T} Q = A P^{T}\right\}$
$=A^{2}P^{T}Q^{2003}P$
$=A^{3}P^{T}Q^{2002}P$
$=A^{2004}P^{T}QP=A^{2004}P^{T}\left(P A\right)$
$=A^{2004}A=A^{2005}$
$\Rightarrow P^{T}Q^{2005}P=A^{2005}=\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}$