1. Tardigrade
  2. Question
  3. Mathematics
  4. If P =[(√3/2)&(1/2) -(1/2)&(√3/2)], A = [1&1 0&1] and Q = PAPT then PT Q2015 P is :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 $P^T \ Q^{2015} P$ is :

JEE MainJEE Main 2016Matrices

Solution:

$P=\begin{vmatrix}\frac{\sqrt{3}}{2}&\frac{1}{2}\\ -\frac{1}{2}&\frac{\sqrt{3}}{2}\end{vmatrix} P^{T}=\begin{vmatrix}\frac{\sqrt{3}}{2}&-\frac{1}{2}\\ \frac{1}{2}&\frac{\sqrt{3}}{2}\end{vmatrix}$
$PP^{T}=P^{T}P=1$
$Q^{2005}=\left(PAP^{T}\right)\left(PAP^{T}\right)........ \, ........\left(2005 \,terms \right)$
$=PA^{2005} P^{T}$
$P^{T}\theta^{2005}P=A^{2005}$
$A^{2}=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}\begin{bmatrix}1&1\\ 0&1\end{bmatrix}=\begin{bmatrix}1&2\\ 0&1\end{bmatrix}$
$A^{3}=\begin{bmatrix}1&2\\ 0&1\end{bmatrix}\begin{bmatrix}1&1\\ 0&1\end{bmatrix}=\begin{bmatrix}1&3\\ 0&1\end{bmatrix}$
$\therefore A^{2005}=\begin{bmatrix}1&2015\\ 0&1\end{bmatrix}$