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Question Number 81430 by abdomathmax last updated on 13/Feb/20

$$letthematrixA=\left(\begin{array}{c}12\\ -13\end{array}\right)$$$$1)calculste{A}^n$$$$2)find{e}^{A}and{e}^{-A}$$$$3)findcosAandsinA$$

Commented by abdomathmax last updated on 13/Feb/20

$$1)P_c\left(A\right)=det(A-xI)=\left|\begin{array}{c}1-x2\\ -13-x\end{array}\right|$$$$=(1-x)(3-x)+2=3-x-3x+x^2+2$$$$=x^2-4x+5\to {\mathrm{\Delta }}^{{}^{\prime }}=4-5=-1\Rightarrow$$$${\lambda }_1=2+iand{\lambda }_2=2-i$$$$x^n=Q\left(x\right)P_c\left(x\right)+u_{n}x+v_n\Rightarrow$$$$A^n=u_{n}A+v_{nI}$$$${\lambda }_1^n=u_n{\lambda }_1+v_{n}and{\lambda }_2^n=u_n{\lambda }_2+v_n\Rightarrow$$$${\lambda }_1^n-{\lambda }_2^n=({\lambda }_1-{\lambda }_2)u_n\Rightarrow u_n=\frac{{\lambda }_1^n-{\lambda }_2^n}{2i}$$$$=\frac{2iIm\left({\lambda }_1^n\right)}{2i}=Im\left({\lambda }_1^n\right)wehave{\lambda }_1=\sqrt{5}{e}^{iarctan\left(\frac{1}{2}\right)}\Rightarrow$$$${\lambda }_1^n={\left(\sqrt{5}\right)}^n{e}^{inarctan\left(\frac{1}{2}\right)}\Rightarrow u_n={\left(\sqrt{5}\right)}^{n}sin\left(narctan\left(\frac{1}{2}\right)\right)$$$$v_n={\lambda }_1^n-{\lambda }_1{u}_n={\lambda }_1^n-{\lambda }_1\times \frac{{\lambda }_1^n-{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}$$$$=\frac{{\lambda }_1^{n+1}-{\lambda }_2{\lambda }_1^n-{\lambda }_1^{n+1}+{\lambda }_1{\lambda }_2^n}{{\lambda }_1-{\lambda }_2}=\frac{{\lambda }_1{\lambda }_2^n-{\lambda }_2{\lambda }_1^n}{2i}$$$$A^n={\left(\sqrt{5}\right)}^{n}sin\left(narctan\left(\frac{1}{2}\right)\right)A+v_{n}I$$$$v_n=\frac{(2+i){(2-i)}^n-(2-i){(2+i)}^n}{2i}$$$$=Im(2+i){(2-i)}^{n}wehave$$$$2+i=\sqrt{5}{e}^{iarctan\left(\frac{1}{2}\right)}and{(2-i)}^n={\left(\sqrt{5}{e}^{-iarctan\left(\frac{1}{2}\right)}\right)}^n$$$$={\left(\sqrt{5}\right)}^n{e}^{-inarctan\left(\frac{1}{2}\right)}\Rightarrow$$$$(2+i){(2-i)}^n={\left(\sqrt{5}\right)}^{n+1}{e}^{-i(n-1)arctan\left(\frac{1}{2}\right)}$$$$\Rightarrow v_n=-{\left(\sqrt{5}\right)}^{n+1}sin(n-1)arctan\left(\frac{1}{2}\right)$$

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