Question

If $P=\left[\begin{array}{cc}5& -3\\ 111& 336\end{array}\right]$ and $det(-3P^{2013}+P^{2014})={\alpha }^{\alpha }{\beta }^2(1+\gamma +{\gamma }^2)$ where $\alpha ,\beta ,\gamma$ are natural numbers, then

• A. $\alpha =2013$
• B. $\beta =3$
• C. $\gamma =10$
• D. $\alpha =2014$

Answer 1

Answer: (A), (B), (C)

$\gamma =10$

$\left|P\right|=\left|\begin{array}{cc}5& -3\\ 111& 336\end{array}\right|=1680+333=2013$

Now, $|-3P^{2013}+P^{2014}|=\left|P^{2013}(-3I+P)\right|={\left|P\right|}^{2013}|-3I+P|$
$={\left(2013\right)}^{2013}\left|\begin{array}{cc}5-3& -3\\ 111& 336-3\end{array}\right|$
$={\left(2013\right)}^{2013}\left|\begin{array}{cc}2& -3\\ 111& 333\end{array}\right|$
$={\left(2013\right)}^{2013}(111\times 9)$
$={\left(2013\right)}^{2013}\cdot 3^2(1+10+{10}^2)$

On comparing, we get
$\alpha =2013,\beta =3$ and $\gamma =10$