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Question Number 99580 by mathmax by abdo last updated on 21/Jun/20

$$\mathrm{let}\mathrm{A}=\left(\begin{array}{c}2-1\\ 31\end{array}\right)$$$$1)\mathrm{prove}\mathrm{that}\mathrm{A}\mathrm{is}\mathrm{inversible}\mathrm{and}\mathrm{calculste}{\mathrm{A}}^{-1}$$$$2)\mathrm{calculate}{\mathrm{A}}^{\mathrm{n}}$$$$3)\mathrm{find}{\mathrm{e}}^{\mathrm{A}}\mathrm{and}{\mathrm{e}}^{-\mathrm{A}}$$$$4)\mathrm{calculate}\cos \mathrm{A}\mathrm{and}\mathrm{sinA}\mathrm{is}{\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}=\mathrm{I}?$$$$$$

Commented by john santu last updated on 22/Jun/20

$$\left(1\right)\det \left(\mathrm{A}\right)\ne 0\Rightarrow {\mathrm{A}}^{-1}=\frac{1}{5}\left[\begin{array}{c}11\\ -32\end{array}\right]$$

Answered by MWSuSon last updated on 22/Jun/20

$$1)\mid \mathrm{A}\mid =2+3=5\ne 0◼$$$$2){\mathrm{A}}^{\mathrm{n}}={\mathrm{PD}}^{\mathrm{n}}{\mathrm{P}}^{-1}$$$$\mid \mathrm{A}-\lambda \mathrm{I}\mid =\left|\begin{array}{cc}2-\lambda & -1\\ 3& 1-\lambda \end{array}\right|=0$$$$(2-\lambda )(1-\lambda )+3=0$$$$\lambda =\frac{1}{2}(3\pm \mathrm{i}\sqrt{11})$$$$\mathrm{when}\lambda =\frac{1}{2}(3+\mathrm{i}\sqrt{11})$$$$\mathrm{let}\mathrm{B}=\left(\begin{array}{cc}\frac{1}{2}-\mathrm{i}\frac{\sqrt{11}}{2}& -1\\ 3& -\frac{1}{2}-\frac{\mathrm{i}\sqrt{11}}{2}\end{array}\right)$$$$\mathrm{B}\overrightarrow{\mathrm{X}}=\left(\begin{array}{cc}\frac{1}{2}-\mathrm{i}\frac{\sqrt{11}}{2}& -1\\ 3& -\frac{1}{2}-\frac{\mathrm{i}\sqrt{11}}{2}\end{array}\right)\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$$$\mathrm{augmented}\mathrm{matrix}$$$$\left(\begin{array}{ccc}\frac{1}{2}-\frac{\mathrm{i}\sqrt{11}}{2}& -1& 0\\ 3& -\frac{1}{2}-\frac{\mathrm{i}\sqrt{11}}{2}& 0\end{array}\right)=\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]$$$$\mathrm{by}\mathrm{gaussian}\mathrm{elimination}$$$$\mathrm{R}2\to \mathrm{R}2-\frac{3}{\frac{1}{2}-\frac{\mathrm{i}\sqrt{11}}{2}}\mathrm{R}1\mathrm{and}\mathrm{R}1\to \frac{2}{1-\mathrm{i}\sqrt{11}}$$$$=\left(\begin{array}{cc}1& -\frac{2}{1-\mathrm{i}\sqrt{11}}\\ 0& 0\end{array}\right)\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$$$${\mathrm{x}}_1-\frac{{2\mathrm{x}}_2}{1-\mathrm{i}\sqrt{11}}=0$$$${\mathrm{x}}_1=\frac{{2\mathrm{x}}_2}{1-\mathrm{i}\sqrt{11}}$$$$\overrightarrow{\mathrm{X}}=\left[\begin{array}{c}\frac{{2\mathrm{ix}}_2}{\mathrm{i}+\sqrt{11}}\\ {\mathrm{x}}_2\end{array}\right],\mathrm{let}{\mathrm{x}}_2=1$$$$\overrightarrow{\mathrm{X}}=\left[\begin{array}{c}\frac{2\mathrm{i}}{\mathrm{i}+\sqrt{11}}\\ 1\end{array}\right]$$$$\mathrm{fellowing}\mathrm{same}\mathrm{procedure}\mathrm{second}$$$$\mathrm{eigenvector}\overrightarrow{\mathrm{X}}=\left[\begin{array}{c}-\frac{2\mathrm{i}}{\sqrt{11}-\mathrm{i}}\\ 1\end{array}\right]$$$$\mathrm{P}=\left[\begin{array}{cc}\frac{2\mathrm{i}}{\sqrt{11}+\mathrm{i}}& -\frac{2\mathrm{i}}{\sqrt{11}-\mathrm{i}}\\ 1& 1\end{array}\right]$$$${\mathrm{P}}^{-1}=\left[\begin{array}{cc}\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}-\frac{\mathrm{i}}{2\sqrt{11}}\\ -\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}+\frac{\mathrm{i}}{2\sqrt{11}}\end{array}\right]$$$$\mathrm{D}={\mathrm{P}}^{-1}\mathrm{AP}$$$$\mathrm{D}=\left[\begin{array}{cc}\frac{1}{2}(3+\mathrm{i}\sqrt{11})& 0\\ 0& \frac{1}{2}(3-\mathrm{i}\sqrt{11})\end{array}\right]$$$$\mathrm{Hence}{\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{cc}\frac{2\mathrm{i}}{\sqrt{11}+\mathrm{i}}& \frac{2\mathrm{i}}{\mathrm{i}-\sqrt{11}}\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{\left(\frac{1}{2}\right)}^{\mathrm{n}}{(3+\mathrm{i}\sqrt{11})}^{\mathrm{n}}& 0\\ 0& {\left(\frac{1}{2}\right)}^{\mathrm{n}}{(3-\sqrt{11})}^{\mathrm{n}}\end{array}\right]\left[\begin{array}{cc}\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}-\frac{\mathrm{i}}{2\sqrt{11}}\\ -\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}+\frac{\mathrm{i}}{2\sqrt{11}}\end{array}\right]$$$$$$

Commented by MWSuSon last updated on 22/Jun/20

you are welcome sir.

Commented by abdomathmax last updated on 22/Jun/20

$$\mathrm{thanks}\mathrm{sir}.$$

Answered by MWSuSon last updated on 22/Jun/20

$$\mathrm{if}\mathrm{A}=\left[\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right]=\left[\begin{array}{cc}2& -1\\ 3& 1\end{array}\right]$$$$\mathrm{then}{\mathrm{e}}^{\mathrm{A}}=\left[\begin{array}{cc}{\mathrm{e}}^{{\mathrm{a}}_{11}}& {\mathrm{e}}^{{\mathrm{a}}_{12}}\\ {\mathrm{e}}^{\mathrm{a}21}& {\mathrm{e}}^{{\mathrm{a}}_{22}}\end{array}\right]=\left[\begin{array}{cc}{\mathrm{e}}^2& \frac{1}{\mathrm{e}}\\ {\mathrm{e}}^3& \mathrm{e}\end{array}\right]$$$${\mathrm{e}}^{-\mathrm{A}}=\left[\begin{array}{cc}\frac{1}{{\mathrm{e}}^2}& \mathrm{e}\\ \frac{1}{{\mathrm{e}}^3}& \frac{1}{\mathrm{e}}\end{array}\right]$$$$4)\mathrm{cosA}=\left[\begin{array}{cc}\cos \left({\mathrm{a}}_{11}\right)& \cos \left({\mathrm{a}}_{12}\right)\\ \cos \left({\mathrm{a}}_{21}\right)& \cos \left({\mathrm{a}}_{22}\right)\end{array}\right]=\left[\begin{array}{cc}\cos 2& \cos 1\\ \cos 3& \cos 1\end{array}\right]$$$$\mathrm{sinA}=\left[\begin{array}{cc}\sin 2& -\sin 1\\ \sin 3& \sin 1\end{array}\right].$$$$\mathrm{yes}{\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}=\mathrm{I}$$

Answered by MWSuSon last updated on 22/Jun/20

$${\mathrm{e}}^{\mathrm{A}}={\mathrm{Pe}}^{\mathrm{D}}{\mathrm{P}}^{-1}$$$${\mathrm{e}}^{-\mathrm{A}}={\mathrm{Pe}}^{-\mathrm{D}}{\mathrm{P}}^{-1}$$$$\Rightarrow {\mathrm{e}}^{\mathrm{A}}=\left[\begin{array}{cc}\frac{2\mathrm{i}}{\sqrt{11}+\mathrm{i}}& \frac{-2\mathrm{i}}{\sqrt{11}-\mathrm{i}}\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{\mathrm{e}}^{\frac{1}{2}(3+\mathrm{i}\sqrt{11})}& 0\\ 0& {\mathrm{e}}^{\frac{1}{2}(3-\mathrm{i}\sqrt{11})}\end{array}\right]\left[\begin{array}{cc}\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}-\frac{\mathrm{i}}{2\sqrt{11}}\\ -\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}+\frac{\mathrm{i}}{2\sqrt{11}}\end{array}\right]$$$$\Rightarrow {\mathrm{e}}^{-\mathrm{A}}=\left[\begin{array}{cc}\frac{2\mathrm{i}}{\sqrt{11}+\mathrm{i}}& \frac{-2\mathrm{i}}{\sqrt{11}-\mathrm{i}}\\ 1& 1\end{array}\right]\left[\begin{array}{cc}{\mathrm{e}}^{-\frac{1}{2}(3+\mathrm{i}\sqrt{11})}& 0\\ 0& {\mathrm{e}}^{-\frac{1}{2}(3-\mathrm{i}\sqrt{11})}\end{array}\right]\left[\begin{array}{cc}\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}-\frac{\mathrm{i}}{2\sqrt{11}}\\ -\frac{3\mathrm{i}}{\sqrt{11}}& \frac{1}{2}+\frac{\mathrm{i}}{2\sqrt{11}}\end{array}\right]$$

Answered by mathmax by abdo last updated on 22/Jun/20

$$1){\mathrm{P}}_{\mathrm{c}}\left(\mathrm{x}\right)=\det (\mathrm{A}-\mathrm{xI})=\left|\begin{array}{c}2-\mathrm{x}-1\\ 31-\mathrm{x}\end{array}\right|=(2-\mathrm{x})(1-\mathrm{x})+3$$$$=(\mathrm{x}-2)(\mathrm{x}-1)+3={\mathrm{x}}^2-3\mathrm{x}+2+3={\mathrm{x}}^2-3\mathrm{x}+5$$$${\mathrm{P}}_{\mathrm{c}}\left(\mathrm{A}\right)=0\Rightarrow {\mathrm{A}}^2-3\mathrm{A}+5\mathrm{I}=0\Rightarrow {\mathrm{A}}^2-3\mathrm{A}=-5\mathrm{I}\Rightarrow$$$$(-\frac{1}{5})\mathrm{A}(\mathrm{A}-3\mathrm{I})=\mathrm{I}\Rightarrow \mathrm{A}\mathrm{is}\mathrm{inversible}\mathrm{and}{\mathrm{A}}^{-1}=\frac{1}{5}(3\mathrm{I}-\mathrm{A})$$$$=\frac{3}{5}\mathrm{I}-\frac{1}{5}\mathrm{A}=\left(\begin{array}{c}\frac{3}{5}0\\ 0\frac{3}{5}\end{array}\right)-\left(\begin{array}{c}\frac{2}{5}-\frac{1}{5}\\ \frac{3}{5}\frac{1}{5}\end{array}\right)=\left(\begin{array}{c}\frac{1}{5}\frac{1}{5}\\ -\frac{3}{5}\frac{2}{5}\end{array}\right)$$

Commented by mathmax by abdo last updated on 22/Jun/20

$$2)\mathrm{we}\mathrm{have}{\mathrm{P}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{x}}^2-3\mathrm{x}+5$$$${\mathrm{P}}_{\mathrm{c}}\left(\mathrm{x}\right)=0\Rightarrow {\mathrm{x}}^2-3\mathrm{x}+5=0\to \mathrm{\Delta }=9-20=-11\Rightarrow {\lambda }_1=\frac{3+\mathrm{i}\sqrt{11}}{2}$$$$\mathrm{and}{\lambda }_2=\frac{3-\mathrm{i}\sqrt{11}}{2}\mathrm{we}\mathrm{divide}{\mathrm{x}}^{\mathrm{n}}\mathrm{by}{\mathrm{p}}_{\mathrm{c}}\left(\mathrm{x}\right)\Rightarrow {\mathrm{x}}^{\mathrm{n}}={\mathrm{qp}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{u}}_{\mathrm{n}}\mathrm{x}+{\mathrm{v}}_{\mathrm{n}}$$$$\Rightarrow {\lambda }_1^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}{\lambda }_1+{\mathrm{v}}_{\mathrm{n}}\mathrm{and}{\lambda }_2^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}{\lambda }_2+{\mathrm{v}}_{\mathrm{n}}\Rightarrow$$$$\frac{{\lambda }_1^{\mathrm{n}}-{\lambda }_2^{\mathrm{n}}}{{\lambda }_1-{\lambda }_2}={\mathrm{u}}_{\mathrm{n}}\Rightarrow {\mathrm{v}}_{\mathrm{n}}={\lambda }_1^{\mathrm{n}}-{\lambda }_1{\mathrm{u}}_{\mathrm{n}}={\lambda }_1^{\mathrm{n}}-{\lambda }_1\times \frac{{\lambda }_1^{\mathrm{n}}-{\lambda }_2^{\mathrm{n}}}{{\lambda }_1-{\lambda }_2}$$$$=\frac{{\lambda }_1^{\mathrm{n}+1}-{\lambda }_2{\lambda }_1^{\mathrm{n}}-{\lambda }_1^{\mathrm{n}+1}+{\lambda }_1{\lambda }_2^{\mathrm{n}}}{{\lambda }_1-{\lambda }_2}=\frac{-5{\lambda }_1^{\mathrm{n}-1}+5{\lambda }_2^{\mathrm{n}-1}}{\mathrm{i}\sqrt{11}}=\frac{-5}{\mathrm{i}\sqrt{11}}({\lambda }_1^{\mathrm{n}-1}-{\lambda }_2^{\mathrm{n}-1})$$$$\mid {\lambda }_1\mid =\frac{1}{2}\left(2\sqrt{5}\right)=\sqrt{5}\Rightarrow {\lambda }_1=\sqrt{5}{\mathrm{e}}^{\mathrm{iarctan}\left(\frac{\sqrt{11}}{3}\right)}\mathrm{and}{\lambda }_2=\sqrt{5}{\mathrm{e}}^{-\mathrm{iarctan}\left(\frac{\sqrt{11}}{3}\right)}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=\frac{{\left(\sqrt{5}\right)}^{\mathrm{n}}\{{\mathrm{e}}^{\mathrm{inarctan}\left(\frac{\sqrt{11}}{3}\right)}-{\mathrm{e}}^{-\mathrm{inarctan}\left(\frac{\sqrt{11}}{3}\right)}\}}{\mathrm{i}\sqrt{11}}=\frac{{\left(\sqrt{5}\right)}^{\mathrm{n}}}{\mathrm{i}\sqrt{11}}\left\{2\mathrm{isin}\left(\mathrm{narctan}\left(\frac{\sqrt{11}}{3}\right)\right)\right\}$$$$=\frac{2{\left(\sqrt{5}\right)}^{\mathrm{n}}}{\sqrt{11}}\sin \left(\mathrm{narctan}\left(\frac{\sqrt{11}}{3}\right)\right)$$$${\mathrm{v}}_{\mathrm{n}}=\frac{-5}{\mathrm{i}\sqrt{11}}\left\{2\mathrm{i}\sin (\mathrm{n}-1)\arctan \left(\frac{\sqrt{11}}{3}\right)\right)\}$$$$=\frac{-10}{\sqrt{11}}\left\{\sin (\mathrm{n}-1)\arctan \left(\frac{\sqrt{11}}{3}\right)\right)\}$$$${\mathrm{A}}^{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}\mathrm{A}+{\mathrm{v}}_{\mathrm{n}}\mathrm{I}={\mathrm{u}}_{\mathrm{n}}\left(\begin{array}{c}2-1\\ 31\end{array}\right)+{\mathrm{v}}_{\mathrm{n}}\left(\begin{array}{c}10\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}{2\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}-{\mathrm{u}}_{\mathrm{n}}\\ {3\mathrm{u}}_{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}+{\mathrm{v}}_{\mathrm{n}}\end{array}\right)$$

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