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 $P^TQ^{2005}P$, is
Answer & explanation
Correct answer: option 1
We have,
$PP^T=\begin{bmatrix}\frac{\sqrt{3}}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{\sqrt{3}}{2}\end{bmatrix}\begin{bmatrix}\frac{\sqrt{3}}{2}&-\frac{1}{2}\\\frac{1}{2}&\frac{\sqrt{3}}{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$
$∴P^T=P^{-1}$
Now,
$Q = PAP^T$
$⇒P^TQP = P^T (PAP^T) P$
$⇒P^TQP =(P^TP)\, A\, (P^TP) = IAI = A =\begin{bmatrix}1&1\\0&1\end{bmatrix}$
and, $Q^2=(PAP^T) (PAP^T) = PA (P^TP) AP^T =PA^2\, P^T$
$∴P^T Q^2 P=P^T (PA^2P^T) P =(P^TP)\, A^2\, (P^TP) = IA^2$
$⇒P^T Q^2 P=A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
Continuing in this manner, we have
$P^TQ^{2005}P=A^{2005}=\begin{bmatrix}1&2005\\0&1\end{bmatrix}$