If $ A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $, find $A^{50}$.

Solution:

$A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, \ |A|=-1$

C.E. is given by,

$|A-\lambda I|=0$

$\begin{vmatrix} 1-\lambda & 0 & 0 \\ 1 & 0-\lambda & 1 \\ 0 & 1 & 0-\lambda \end{vmatrix}=0$

$\lambda^{3}-(1)\lambda^{2}+(0+0+0-0-0-1)\lambda+1=0$

$\lambda^{3}-\lambda^{2}-\lambda+1=0$

$\therefore …