Question

If $M=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$ and $M^{2}-\lambda M-I_{2}=O$ , then $2^{\lambda }$ must be

Answer 1

We have,
$M^{2}=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}=\begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$
$\lambda M=\begin{bmatrix} \lambda & 2\lambda \\ 2\lambda & 3\lambda \end{bmatrix}$
$\therefore M^{2}-\lambda M-l_{2}=0$
$\Rightarrow \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}-\begin{bmatrix} \lambda & 2\lambda \\ 2\lambda & 3\lambda \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 4-\lambda & 8-2\lambda \\ 8-2\lambda & 12-3\lambda \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\therefore \lambda =4$
$\therefore 2^{\lambda }=2^{4}=16$