JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q64

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If $P=\begin{bmatrix}\frac{\sqrt3}2&\frac12\\-\frac12&\frac{\sqrt3}2\end{bmatrix}$ , $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $Q=PAP^T$ , then $P^TQ^{2015}P$ is :

\begin{bmatrix}0&2015\\0&0\end{bmatrix}

\begin{bmatrix}2015&1\\0&2015\end{bmatrix}

\begin{bmatrix}2015&0\\1&2015\end{bmatrix}

\begin{bmatrix}1&2015\\0&1\end{bmatrix}

Answer and worked solution

Official answer: D

P is a rotation, so $P^TP=I$ and $Q^{2015}=PA^{2015}P^T$ . Then $P^TQ^{2015}P=A^{2015}=\begin{bmatrix}1&2015\\0&1\end{bmatrix}$ .

Topics:Matrix algebra Special matrices

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 64