written 4.9 years ago by | • modified 4.9 years ago |
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$\begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\ 0& 0 & 1 \end{bmatrix}$
$A^{50}=\begin{bmatrix} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25& 0 & 1 \end{bmatrix}$