Question

Let $A=\left[\begin{array}{cc}\tan \frac{\pi }{3}& \sec \frac{2\pi }{3}\\ \cot \left(2013\frac{\pi }{2}\right)& \cos (2012\pi )\end{array}\right]$
and $P$ is a $2\times 2$ matrix such that $P{P}^T=I$, where $I$ is an identity matrix of order $2$ . If $Q=PA{P}^T$ and $R={\left[r_{ij}\right]}_{2\times 2}=P^T{Q}^{8}P$, then

• A. $r_{11}=81$
• B. $r_{11}=27\sqrt{3}$
• C. $r_{11}=4\sqrt{3}$
• D. $r_{11}=-\sqrt{3}$

Answer 1

Answer: (A)

$R=P^T{Q}^{8}P=A^8$
Now, $A^2=AA=\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& -2\sqrt{3}-2\\ 0& 1\end{array}\right]$
Also, $A^3=A^{2}A$
$=\left[\begin{array}{cc}3& -2\sqrt{3}-2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}(\sqrt{3}{)}^3& -6-2\sqrt{3}-2\\ 0& 1\end{array}\right]$
$\therefore R={\left[r_{ij}\right]}_{2\times 2}$
$=P^T{Q}^{8}P=A^8=\left[\begin{array}{cc}(\sqrt{3}{)}^8& -\\ -& -\end{array}\right]$
$\Rightarrow r_{11}=81$