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,

$|A-\lambda I|=0$

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

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

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

$\therefore \lambda=-1,1,1$

Eigen values are repeated.

As it is a 3x3 matrix,

$\phi(A)=A^{50}= \alpha A^{2}+ \beta A + \gamma I$ -------------(1)

$\lambda$ satisfies the above equation,

$\lambda^{50} = \alpha \lambda^{2}+ \beta \lambda +\gamma$ ------------(2)

Put $\lambda=1$

$(1)^{50}= \alpha (1)^{2}+\beta (1) + \gamma$

$1=\alpha+\beta+\gamma$ ---------(3)

Put $\lambda=-1$

$(-1)^{50}= \alpha (-1)^{2}+\beta (-1) + \gamma$

$1=\alpha - \beta+\gamma$ ---------------(4)

Since eigen values are repeated differentiating (2) w.r.t. $\lambda$

$50 \lambda^{49}= 2 \alpha \lambda + \beta +0$

$50 \lambda^{49}= 2 \alpha \lambda + \beta$

Put $\lambda=1$ in the above equation,

$50 (1)^{49}= 2 \alpha (1) + \beta$

$50=2 \alpha +\beta$ ---------(5)

Solve equation (3), (4) and (5) simulatneously, we get

$\alpha = 25, \ \beta=0, \ \gamma=-24$

Put $\alpha, \ \beta , \ \gamma $ values in equation (1),

$A^{50}=25 A^{2}+0.A+(-24)I$

$A^{50}=25A^{2}-24I$

$A^{50}=25\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}-24\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$A^{50}=\begin{bmatrix} 25 & 0 & 0 \\ 25 & 25 & 0 \\ 25 & 0 & 25 \end{bmatrix}-24\begin{bmatrix} 24 & 0 & 0 \\ 0 & 24 & 0 \\ 0 & 0 & 24 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{bmatrix}$