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Question Number 122231 by fajri last updated on 15/Nov/20

Answered by mathmax by abdo last updated on 15/Nov/20

$${\mathrm{e}}^{\mathrm{tA}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}{\mathrm{A}}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{let}\mathrm{determine}{\mathrm{A}}^{\mathrm{n}}$$$${\mathrm{P}}_{\mathrm{c}}\left(\mathrm{A}\right)=\det (\mathrm{A}-\mathrm{xI})=\left|\begin{array}{c}3-\mathrm{x}-11\\ 2-\mathrm{x}1\end{array}\right|$$$$\mid 1-12-\mathrm{x}\mid$$$$=(3-\mathrm{x})\left|\begin{array}{c}-\mathrm{x}1\\ -12-\mathrm{x}\end{array}\right|-2\left|\begin{array}{c}-11\\ -12-\mathrm{x}\end{array}\right|+\left|\begin{array}{c}-11\\ -\mathrm{x}1\end{array}\right|$$$$=(3-\mathrm{x})(-\mathrm{x}(2-\mathrm{x})+1)-2(-2+\mathrm{x}+1)-1+\mathrm{x}$$$$=(3-\mathrm{x})(-2\mathrm{x}+{\mathrm{x}}^2+1)-2(\mathrm{x}-1)+(\mathrm{x}-1)$$$$={(\mathrm{x}-1)}^2(3-\mathrm{x})-(\mathrm{x}-1)=(\mathrm{x}-1)((\mathrm{x}-1)(3-\mathrm{x})-1)$$$$=(\mathrm{x}-1)(3\mathrm{x}-{\mathrm{x}}^2-3+\mathrm{x}-1)=(\mathrm{x}-1)(-{\mathrm{x}}^2+4\mathrm{x}-1)$$$$=(1-\mathrm{x})({\mathrm{x}}^2-4\mathrm{x}+1)$$$${\mathrm{x}}^2-4\mathrm{x}+1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-1=3\Rightarrow {\mathrm{x}}_1=2+\sqrt{3}\mathrm{and}{\mathrm{x}}_2=2-\sqrt{3}$$$$\Rightarrow {\lambda }_1=1,{\lambda }_2=2+\sqrt{3},{\lambda }_3=2-\sqrt{3}$$$$\mathrm{if}\mathrm{we}\mathrm{divide}{\mathrm{x}}^{\mathrm{n}}\mathrm{by}{\mathrm{p}}_{\mathrm{c}}\left(\mathrm{A}\right)\mathrm{we}\mathrm{get}{\mathrm{x}}^{\mathrm{n}}={\mathrm{qp}}_{\mathrm{c}}\left(\mathrm{A}\right)+{\mathrm{u}}_{\mathrm{n}}{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{n}}\mathrm{x}+{\mathrm{s}}_{\mathrm{n}}$$$$1={\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}+{\mathrm{s}}_{\mathrm{n}}\mathrm{snd}{(2+\sqrt{3})}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}{(2+\sqrt{3})}^2+{\mathrm{v}}_{\mathrm{n}}(2+\sqrt{3})+{\mathrm{s}}_{\mathrm{n}}$$$${(2-\sqrt{3})}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}{(2-\sqrt{3})}^2+{\mathrm{v}}_{\mathrm{n}}(2-\sqrt{3})+{\mathrm{s}}_{\mathrm{n}}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{system}$$$$\{\begin{array}{l}{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}+{\mathrm{s}}_{\mathrm{n}}=1\\ {(2+\sqrt{3})}^2{\mathrm{u}}_{\mathrm{n}}+(2+\sqrt{3}){\mathrm{v}}_{\mathrm{n}}+{\mathrm{s}}_{\mathrm{n}}={(2+\sqrt{3})}^{\mathrm{n}}\end{array}$$$$\{{(2-\sqrt{3})}^2{\mathrm{u}}_{\mathrm{n}}+(2-\sqrt{3}){\mathrm{v}}_{\mathrm{n}}+{\mathrm{s}}_{\mathrm{n}}={(2-\sqrt{3})}^{\mathrm{n}}\Rightarrow$$$$\left(\begin{array}{c}111\\ {(2+\sqrt{3})}^{2}2+\sqrt{3}1\end{array}\right).\left(\begin{array}{c}{\mathrm{u}}_{\mathrm{n}}\\ {\mathrm{v}}_{\mathrm{n}}\end{array}\right)=\left(\begin{array}{c}1\\ {(2+\sqrt{3})}^{\mathrm{n}}\end{array}\right)$$$$({(2-\sqrt{3})}^{2}2-\sqrt{3}1)\left({\mathrm{s}}_{\mathrm{n}}\right)\left({(2-\sqrt{3})}^{\mathrm{n}}\right)\Rightarrow$$$$\mathrm{M}.\left(\begin{array}{c}{\mathrm{u}}_{\mathrm{n}}\\ {\mathrm{v}}_{\mathrm{n}}\end{array}\right)=\left(\begin{array}{c}1\\ {(2+\sqrt{3})}^{\mathrm{n}}\end{array}\right)$$$$\left({\mathrm{s}}_{\mathrm{n}}\right)\left({(2-\sqrt{3})}^{\mathrm{n}}\right)\Rightarrow$$$$\left(\begin{array}{c}{\mathrm{u}}_{\mathrm{n}}\\ {\mathrm{v}}_{\mathrm{n}}\end{array}\right)={\mathrm{M}}^{-1}\left(\begin{array}{c}1\\ {(2+\sqrt{3})}^{\mathrm{n}}\end{array}\right)$$$$\left({\mathrm{s}}_{\mathrm{n}}\right)\left((2-\sqrt{3})\right)\mathrm{we}\mathrm{have}$$$${\mathrm{A}}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}{\mathrm{A}}^2+{\mathrm{v}}_{\mathrm{n}}\mathrm{A}+{\mathrm{s}}_{\mathrm{n}}\mathrm{I}\mathrm{rest}\mathrm{to}\mathrm{find}{\mathrm{M}}^{-1}=\frac{{\mathrm{t}}_{\mathrm{com}\left(\mathrm{M}\right)}}{\det \mathrm{M}}$$$$...\mathrm{be}\mathrm{continued}....$$

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