Question

If $M=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ and $M^2-\lambda M-I_2=O$ , then $2^{\lambda }$ must be

Answer 1

Answer: 16

We have,
$M^2=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]$
$\lambda M=\left[\begin{array}{cc}\lambda & 2\lambda \\ 2\lambda & 3\lambda \end{array}\right]$
$\therefore M^2-\lambda M-l_2=0$
$\Rightarrow \left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-\left[\begin{array}{cc}\lambda & 2\lambda \\ 2\lambda & 3\lambda \end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}4-\lambda & 8-2\lambda \\ 8-2\lambda & 12-3\lambda \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\therefore \lambda =4$
$\therefore 2^{\lambda }=2^4=16$