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Question Number 99828 by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{let}\mathrm{A}=\left(\begin{array}{c}21\\ 12\end{array}\right)$$$$1)\mathrm{calculate}{\mathrm{A}}^{\mathrm{n}}$$$$2)\mathrm{determine}\mathrm{cosA}\mathrm{and}\mathrm{sinA}$$$$3)\mathrm{find}\mathrm{chA}\mathrm{and}\mathrm{shA}$$

Commented by bachamohamed last updated on 23/Jun/20

$$1)$$$$\mathit{A}=\left(\begin{array}{c}21\\ 12\end{array}\right)$$$${\mathit{p}}_{\mathit{A}}\left(\mathit{\lambda }\right)=\mathit{d}\mathit{e}\mathit{t}(\mathit{A}-\mathit{\lambda }{\mathit{I}}_2)=(\mathit{\lambda }-1)(\mathit{\lambda }-2)\mathit{d}\mathit{o}\mathit{n}\mathit{c}\mathit{l}\mathit{e}\mathit{s}\mathit{v}\mathit{a}\mathit{l}\mathit{e}\mathit{u}\mathit{r}\mathit{s}\mathit{p}\mathit{r}\mathit{o}\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{o}\mathit{n}\mathit{t}\{{}_{{\mathit{\lambda }}_2=2}^{{\mathit{\lambda }}_1=1}$$$${\mathit{E}}_{{\mathit{\lambda }}_1}=<(-1.1)>{\mathit{E}}_{{\mathit{\lambda }}_2}=<(1.1)>\mathit{o}\mathit{u}\mathit{l}\mathit{e}\mathit{s}\mathit{v}\mathit{e}\mathit{c}\mathit{t}\mathit{e}\mathit{u}\mathit{r}\mathit{s}\mathit{p}\mathit{r}\mathit{o}\mathit{p}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{o}\mathit{n}\mathit{t}\{{}_{{\mathit{v}}_2=(1.1)}^{{\mathit{v}}_1=(-1.1)}$$$$\mathit{e}\mathit{t}\mathit{d}\mathit{o}\mathit{n}\mathit{c}\mathrm{\exists }\mathit{p}\in {\mathit{M}}_{2\mathrm{x}2}\left(\mathrm{R}\right)\mathit{t}\mathit{e}\mathit{l}\mathit{l}\mathit{e}\mathit{q}\mathit{u}\mathit{e}\mathit{B}={\mathit{p}}^{-1}\mathit{A}.\mathit{p}$$$$\mathit{o}\mathit{n}\mathit{a}{\mathit{i}\mathit{d}}_{\mathit{E}}({\mathit{R}}^2.({\mathit{v}}_1.{\mathit{v}}_2))\stackrel{\mathit{p}\mathit{m}\mathit{a}\mathit{t}\mathit{r}\mathit{i}\mathit{x}\mathit{d}\mathit{e}{\mathit{i}\mathit{d}}_{\mathit{E}}}{\to }({\mathit{R}}^2.({\mathit{e}}_1.{\mathit{e}}_2))\mathit{e}\mathit{n}\mathit{d}\mathit{e}\mathit{d}\mathit{u}\mathit{i}\mathit{r}\mathit{e}\mathit{q}\mathit{u}\mathit{e}\mathit{p}=\left(\begin{array}{c}-11\\ 11\end{array}\right)\mathit{e}\mathit{t}{\mathit{p}}^{-1}=\left(\begin{array}{c}-\frac{1}{2}\frac{1}{2}\\ \frac{1}{2}\frac{1}{2}\end{array}\right)$$$$\mathit{B}={\mathit{p}}^{-1}\mathit{A}\mathit{p}=\left(\begin{array}{c}10\\ 03\end{array}\right)\mathit{e}\mathit{t}{\mathit{B}}^{\mathit{n}}={\left({\mathit{p}}^{-1}\mathit{A}\mathit{p}\right)}^{\mathit{n}}=\left({\mathit{p}}^{-1}\mathit{A}\mathit{p}\right)\left({\mathit{p}}^{-1}\mathit{A}\mathit{p}\right).....\stackrel{\mathit{n}\mathit{f}\mathit{o}\mathit{i}\mathit{s}}{.}......\left({\mathit{p}}^{-1}\mathit{A}\mathit{p}\right)={\mathit{p}}^{-1}{\mathit{A}}^{\mathit{n}}.\mathit{p}\Rightarrow {\mathit{A}}^{\mathit{n}}=\mathit{p}.{\mathit{B}}^{\mathit{n}}.{\mathit{p}}^{-1}$$$$\Rightarrow {\mathit{A}}^{\mathit{n}}=\frac{1}{2}\left(\begin{array}{c}(3^{\mathit{n}}+1)3^{\mathit{n}}-1\\ 3^{\mathit{n}}-13^{\mathit{n}}+1\end{array}\right)$$

Commented by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{thankx}\mathrm{sir}.$$

Answered by mathmax by abdo last updated on 23/Jun/20

$$1)\mathrm{the}\mathrm{caracteristic}\mathrm{polynome}\mathrm{is}{\mathrm{P}}_{\mathrm{c}}\left(\mathrm{A}\right)=\det (\mathrm{A}-\mathrm{xI})=\left|\begin{array}{c}2-\mathrm{x}1\\ 12-\mathrm{x}\end{array}\right|$$$$={(2-\mathrm{x})}^2-1={(\mathrm{x}-2)}^2-1=(\mathrm{x}-2-1)(\mathrm{x}-2+1)=(\mathrm{x}-3)(\mathrm{x}-1)\Rightarrow$$$$\Rightarrow {\lambda }_1=1\mathrm{and}{\lambda }_2=3\mathrm{we}\mathrm{have}{\mathrm{x}}^{\mathrm{n}}={\mathrm{p}}_{\mathrm{c}}\left(\mathrm{x}\right)\mathrm{q}+{\mathrm{u}}_{\mathrm{n}}\mathrm{x}+{\mathrm{v}}_{\mathrm{n}}\Rightarrow$$$$1={\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}\mathrm{and}{3}^{\mathrm{n}}={3\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}\Rightarrow {2\mathrm{u}}_{\mathrm{n}}=3^{\mathrm{n}}-1\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{3^{\mathrm{n}}-1}{2}$$$${\mathrm{v}}_{\mathrm{n}}=1-{\mathrm{u}}_{\mathrm{n}}=1-\frac{3^{\mathrm{n}}-1}{2}=\frac{2-3^{\mathrm{n}}+1}{2}=\frac{3-3^{\mathrm{n}}}{2}$$$$\mathrm{we}\mathrm{have}{\mathrm{A}}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}\mathrm{A}+{\mathrm{v}}_{\mathrm{n}}\mathrm{I}=\frac{3^{\mathrm{n}}-1}{2}\left(\begin{array}{c}21\\ 12\end{array}\right)+\frac{3-3^{\mathrm{n}}}{2}\left(\begin{array}{c}10\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}{3}^{\mathrm{n}}-1\frac{3^{\mathrm{n}}-1}{2}\\ \frac{3^{\mathrm{n}}-1}{2}{3}^{\mathrm{n}}-1\end{array}\right)+\left(\begin{array}{c}\frac{3-3^{\mathrm{n}}}{2}0\\ 0\frac{3-3^{\mathrm{n}}}{2}\end{array}\right)$$$$=\left(\begin{array}{c}\frac{2.3^{\mathrm{n}}-2+3-3^{\mathrm{n}}}{2}\frac{3^{\mathrm{n}}-1}{2}\\ \frac{3^{\mathrm{n}}-1}{2}\frac{2.3^{\mathrm{n}}-2+3-3^{\mathrm{n}}}{2}\end{array}\right)$$$$=\left(\begin{array}{c}\frac{3^{\mathrm{n}}+1}{2}\frac{3^{\mathrm{n}}-1}{2}\\ \frac{3^{\mathrm{n}}-1}{2}\frac{3^{\mathrm{n}}+1}{2}\end{array}\right)$$

Commented by mathmax by abdo last updated on 24/Jun/20

$$2)\cos \mathrm{A}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{A}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\left(2\mathrm{n}\right)!}\left(\begin{array}{c}{3}^{2\mathrm{n}}+13^{2\mathrm{n}}-1\\ 3^{2\mathrm{n}}-13^{2\mathrm{n}}+1\end{array}\right)$$$$=\frac{1}{2}\left(\begin{array}{c}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{3}^{2\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{3}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}\\ .........\end{array}\right)$$$$=\frac{1}{2}\left(\begin{array}{c}\cos \left(3\right)+\cos \left(1\right)\cos 3-\cos \left(1\right)\\ \cos \left(3\right)-\cos \left(1\right)\cos \left(3\right)+\cos \left(1\right)\end{array}\right)$$$$\mathrm{sinA}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{A}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}=.....$$

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