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Question Number 11035 by suci last updated on 08/Mar/17

$$A=\left[\begin{array}{cc}1& -1\\ -4& -2\end{array}\right]$$$$A^{2016}=.....?$$

Answered by diofanto last updated on 19/Jul/17

$$\mathrm{Finding}\mathrm{the}\mathrm{eigenvalues}\mathrm{of}A:$$$$\left|\begin{array}{cc}1-\lambda & -1\\ -4& -2-\lambda \end{array}\right|=0$$$$(1-\lambda )(-2-\lambda )-4=0$$$${\lambda }^2+\lambda -6=0$$$${\lambda }_1=2,{\lambda }_2=-3$$$$\mathrm{Finding}\mathrm{an}\mathrm{eigenvector}\mathrm{for}\mathrm{each}\mathrm{eigenvalue}:$$$$Ax={\lambda }_{1}x\iff (A-{\lambda }_{1}I)x=0$$$$\iff \left[\begin{array}{cc}-1& -1\\ -4& -4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$$$\iff x_1=-x_2,\mathrm{e}.\mathrm{g}.\left[\begin{array}{c}-1\\ 1\end{array}\right]$$$$Ay={\lambda }_{2}y\iff (A-{\lambda }_{2}I)y=0$$$$\iff \left[\begin{array}{cc}4& -1\\ -4& 1\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$$$\iff y_1=y_2/4,\mathrm{e}.\mathrm{g}.\left[\begin{array}{c}1\\ 4\end{array}\right]$$$$\mathrm{Diagonalizing}:$$$$A=\left[\begin{array}{cc}-1& 1\\ 1& 4\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& -3\end{array}\right]{\left[\begin{array}{cc}-1& 1\\ 1& 4\end{array}\right]}^{-1}$$$$\mathrm{Finding}{\left[\begin{array}{cc}-1& 1\\ 1& 4\end{array}\right]}^{-1}:$$$$\left(\begin{array}{cccc}-1& 1& 1& 0\\ 1& 4& 0& 1\end{array}\right)\sim \left(\begin{array}{cccc}-1& 1& 1& 0\\ 0& 5& 1& 1\end{array}\right)$$$$\sim \left(\begin{array}{cccc}-1& 0& 4/5& -1/5\\ 0& 5& 1& 1\end{array}\right)\sim \left(\begin{array}{cccc}1& 0& -4/5& 1/5\\ 0& 1& 1/5& 1/5\end{array}\right)$$$$\iff {\left[\begin{array}{cc}-1& 1\\ 1& 4\end{array}\right]}^{-1}=\left[\begin{array}{cc}-4/5& 1/5\\ 1/5& 1/5\end{array}\right]$$$$A^{2016}=\left[\begin{array}{cc}-1& 1\\ 1& 4\end{array}\right]\left[\begin{array}{cc}{2}^{2016}& 0\\ 0& {(-3)}^{2016}\end{array}\right]\left[\begin{array}{cc}-4/5& 1/5\\ 1/5& 1/5\end{array}\right]$$$$\dots$$

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