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Question Number 130771 by mathmax by abdo last updated on 28/Jan/21

$$\mathrm{let}\mathrm{A}=\left(\begin{array}{c}12\\ 21\end{array}\right)$$$$1)\mathrm{find}{\mathrm{e}}^{\mathrm{A}}\mathrm{and}{\mathrm{e}}^{-\mathrm{A}}$$$$2)\mathrm{calculate}\mathrm{ch}\left(\mathrm{A}\right)\mathrm{and}\mathrm{sh}\left(\mathrm{A}\right)$$

Answered by Olaf last updated on 28/Jan/21

$$\mathrm{A}=\left(\begin{array}{c}12\\ 21\end{array}\right)$$$${\mathrm{A}}^2=\left(\begin{array}{c}54\\ 45\end{array}\right)=2\mathrm{A}+3\mathrm{I}$$$${\mathrm{A}}^3={\mathrm{AA}}^2=\mathrm{A}(2\mathrm{A}+3\mathrm{I})=7\mathrm{A}+6\mathrm{I}$$$$...$$$$\mathrm{Let}{\mathrm{A}}^n=a_n\mathrm{A}+b_n\mathrm{I},a_1=1,b_1=0$$$$\Rightarrow {\mathrm{A}}^{n+1}={\mathrm{AA}}^n=\mathrm{A}(a_n\mathrm{A}+b_n\mathrm{I})$$$${\mathrm{A}}^{n+1}=a_n{\mathrm{A}}^2+b_n\mathrm{A}$$$${\mathrm{A}}^{n+1}=(2a_n+b_n)\mathrm{A}+3a_n\mathrm{I}$$$$\Rightarrow \{\begin{array}{l}{a}_{n+1}=2a_n+b_n\left(1\right)\\ b_{n+1}=3a_n\left(2\right)\end{array}$$$$\left(1\right):a_{n+2}=2a_{n+1}+b_{n+1}$$$$a_{n+2}=2a_{n+1}+3a_n,a_1=1,a_2=2$$$$\Rightarrow a_n=\frac{1}{4}[3^n+{(-1)}^{n+1}]$$$$\left(2\right):b_n=3a_{n-1}=\frac{1}{4}[3^n+3{(-1)}^n]$$$${\mathrm{A}}^n=a_n\mathrm{A}+b_n\mathrm{I}$$$${\mathrm{A}}^n=\frac{1}{2}\left(\begin{array}{cc}{3}^n+{(-1)}^n& 3^n+{(-1)}^{n+1}\\ 3^n+{(-1)}^{n+1}& 3^n+{(-1)}^n\end{array}\right)$$$${\mathrm{A}}^{-n}=\frac{1}{2}\left(\begin{array}{cc}\frac{1}{3^n}+{(-1)}^n& \frac{1}{3^n}+{(-1)}^n\\ \frac{1}{3^n}+{(-1)}^n& \frac{1}{3^n}+{(-1)}^n\end{array}\right)$$$$e^{\mathrm{A}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{\mathrm{A}}^n$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{3^n}{n!}=e^3,\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^{n}n!}=\frac{1}{e^3},\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}=\frac{1}{e}$$$$e^{\mathrm{A}}=\frac{1}{2}\left(\begin{array}{cc}{e}^3+\frac{1}{e}& e^3-\frac{1}{e}\\ e^3-\frac{1}{e}& e^3+\frac{1}{e}\end{array}\right)$$$$e^{-\mathrm{A}}=\frac{1}{2}\left(\begin{array}{cc}\frac{1}{e^3}+\frac{1}{e}& \frac{1}{e^3}+\frac{1}{e}\\ \frac{1}{e^3}+\frac{1}{e}& \frac{1}{e^3}+\frac{1}{e}\end{array}\right)$$$$\mathrm{coshA}=\frac{1}{2}(e^{\mathrm{A}}+e^{-\mathrm{A}})$$$$\mathrm{coshA}=\frac{1}{4}\left(\begin{array}{cc}{e}^3+\frac{1}{e^3}+\frac{2}{e}& e^3+\frac{1}{e^3}\\ e^3+\frac{1}{e^3}& e^3+\frac{1}{e^3}+\frac{2}{e}\end{array}\right)$$$$\mathrm{sinhA}=\frac{1}{2}(e^{\mathrm{A}}-e^{-\mathrm{A}})$$$$\mathrm{sinhA}=\frac{1}{4}\left(\begin{array}{cc}{e}^3-\frac{1}{e^3}& e^3-\frac{1}{e^3}-\frac{2}{e}\\ e^3-\frac{1}{e^3}-\frac{2}{e}& e^3-\frac{1}{e^3}\end{array}\right)$$

Commented by mathmax by abdo last updated on 29/Jan/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{Olaf}$$

Answered by mathmax by abdo last updated on 29/Jan/21

$$1){\mathrm{p}}_{\mathrm{c}}\left(\mathrm{A}\right)=\det (\mathrm{A}-\mathrm{xI})=\left|\begin{array}{c}1-\mathrm{x}2\\ 21-\mathrm{x}\end{array}\right|={(1-\mathrm{x})}^2-4$$$$=(1-\mathrm{x}-2)(1-\mathrm{x}+2)=(-\mathrm{x}-1)(3-\mathrm{x})=(\mathrm{x}+1)(\mathrm{x}-3)$$$$\Rightarrow {\lambda }_1=-1\mathrm{and}{\lambda }_2=3$$$${\mathrm{x}}^{\mathrm{n}}=\mathrm{q}\left(\mathrm{x}\right)\mathrm{p}\left(\mathrm{x}\right)+{\mathrm{u}}_{\mathrm{n}}\mathrm{x}+{\mathrm{v}}_{\mathrm{n}}\mathrm{we}\mathrm{get}{(-1)}^{\mathrm{n}}=-{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}\mathrm{and}{3}^{\mathrm{n}}={3\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}\Rightarrow$$$$\{\begin{array}{l}-{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}={(-1)}^{\mathrm{n}}\\ {3\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}=3^{\mathrm{n}}\Rightarrow \{\begin{array}{l}{4\mathrm{u}}_{\mathrm{n}}=3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}\Rightarrow \{\begin{array}{l}{\mathrm{u}}_{\mathrm{n}}=\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{4}\\ {\mathrm{v}}_{\mathrm{n}}=\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{4}+{(-1)}^{\mathrm{n}}\end{array}\\ {\mathrm{v}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+{(-1)}^{\mathrm{n}}\end{array}\end{array}$$$$\Rightarrow \{\begin{array}{l}{\mathrm{u}}_{\mathrm{n}}=\frac{3^{\mathrm{n}}-{(-1)}^{\mathbf{n}}}{4}\\ {\mathbf{v}}_{\mathbf{n}}=\frac{3^{\mathbf{n}}+3{(-1)}^{\mathbf{n}}}{4}\end{array}$$$${\mathbf{A}}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}\mathrm{A}+{\mathrm{v}}_{\mathrm{n}}\mathrm{I}=\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{4}\left(\begin{array}{c}12\\ 21\end{array}\right)+\frac{3^{\mathrm{n}}+3{(-1)}^{\mathrm{n}}}{4}\left(\begin{array}{c}10\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{4}+\frac{3^{\mathrm{n}}+3{(-1)}^{\mathrm{n}}}{4}\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\\ \frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{4}+\frac{3^{\mathrm{n}}+3{(-1)}^{\mathrm{n}}}{4}\end{array}\right)$$$$=\left(\begin{array}{c}\frac{2.3^{\mathrm{n}}+2{(-1)}^{\mathrm{n}}}{4}\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\\ \frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\frac{2.3^{\mathrm{n}}+2{(-1)}^{\mathrm{n}}}{4}\end{array}\right)$$$$=\left(\begin{array}{c}\frac{3^{\mathrm{n}}+{(-1)}^{\mathrm{n}}}{2}\frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\\ \frac{3^{\mathrm{n}}-{(-1)}^{\mathrm{n}}}{2}\frac{3^{\mathrm{n}}+{(-1)}^{\mathrm{n}}}{2}\end{array}\right)$$$${\mathrm{e}}^{\mathrm{A}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{A}}^{\mathrm{n}}}{\mathrm{n}!}=\frac{1}{2}\left(\begin{array}{c}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{3^{\mathrm{n}}}{\mathrm{n}!}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{3^{\mathrm{n}}}{\mathrm{n}!}-\sum _{\mathrm{n}0}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}\\ (.....).....\end{array}\right)$$$$=\frac{1}{2}\left(\begin{array}{c}{\mathrm{e}}^3+{\mathrm{e}}^{-1}{\mathrm{e}}^3-{\mathrm{e}}^{-1}\\ {\mathrm{e}}^3-{\mathrm{e}}^{-1}{\mathrm{e}}^3+{\mathrm{e}}^{-1}\end{array}\right)$$$$$$

Commented by mathmax by abdo last updated on 29/Jan/21

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

Commented by mathmax by abdo last updated on 29/Jan/21

$$\mathrm{ch}\left(\mathrm{A}\right)=\frac{1}{2}({\mathrm{e}}^{\mathrm{A}}+{\mathrm{e}}^{-\mathrm{A}})=\frac{1}{4}\left(\begin{array}{c}{\mathrm{e}}^3+{\mathrm{e}}^{-1}{\mathrm{e}}^3-{\mathrm{e}}^{-1}\\ {\mathrm{e}}^3-{\mathrm{e}}^{-1}{\mathrm{e}}^3+{\mathrm{e}}^{-1}\end{array}\right)$$$$+\frac{1}{4}\left(\begin{array}{c}{\mathrm{e}}^{-3}+\mathrm{e}{\mathrm{e}}^{-3}-\mathrm{e}\\ {\mathrm{e}}^{-3}-\mathrm{e}{\mathrm{e}}^{-3}+\mathrm{e}\end{array}\right)$$$$=\frac{1}{4}\left(\begin{array}{c}{\mathrm{e}}^3+{\mathrm{e}}^{-1}+{\mathrm{e}}^{-3}+\mathrm{e}{\mathrm{e}}^3-{\mathrm{e}}^{-1}+{\mathrm{e}}^{-3}-\mathrm{e}\\ {\mathrm{e}}^3-{\mathrm{e}}^{-1}+{\mathrm{e}}^{-3}-\mathrm{e}{\mathrm{e}}^3+{\mathrm{e}}^{-1}+{\mathrm{e}}^{-3}+\mathrm{e}\end{array}\right)$$$$$$

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