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$A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0& 1 & 0 \end{bmatrix} find A^{50}$
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Solution:

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

characteristic equation is given by

$|A-\lambda I|=0$

$\left |\begin{matrix} 1-\lambda & 0 & 0 \\ 1 & 0-\lambda & 1 \\ 0& 1 & 0-\lambda \end{matrix} \right |$=0

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

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

$\lambda =-1,1,1$

Eigen values are repeated.

Since the matrix is of order3,

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

$\lambda$ satisfies this equation

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

Put $\lambda = -1$ in eqn(2)

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

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

put $\lambda =1$ in (2)

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

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

diff.(2) w.r.t. $\lambda$

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

$50\lambda^{49} = 2 \alpha \lambda + \beta$-----(5)

Put $\lambda = 1$ in eqn(5)

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

$50=2\alpha +\beta$----(6)

solve eqn (3),(4),(5) simultaneously, we get

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

Put $\alpha \ and \ \beta$ values in eqn(1)

$A^{50} = 25A^{2}-24I$-----(7)

$A^{2}=A.A=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0& 1 & 0 \end{bmatrix}$ $\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0& 1 & 0 \end{bmatrix}$

$A^2=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1& 1 \\ 1& 0 & 1 \end{bmatrix}$----(8)

$I= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 1 \end{bmatrix}$----(9)

Put eqn(8),(9) in eqn (7)

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

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

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