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Question Number 107547 by hgrocks last updated on 11/Aug/20

Commented by hgrocks last updated on 11/Aug/20

Can anyone solve this Q without using eigen values

Commented by bemath last updated on 11/Aug/20

$$tr\left(M\right)=trace\left(M\right)?$$

Commented by hgrocks last updated on 11/Aug/20

Yes it is trace of M

Commented by hgrocks last updated on 11/Aug/20

$$\mathrm{Pls}\mathrm{Solve}\mathrm{This}\mathrm{Q}$$

Commented by hgrocks last updated on 11/Aug/20

$$\mathrm{Anyone}?$$$$$$$$$$

Commented by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{sir}\mathrm{dont}\mathrm{beg}\mathrm{the}\mathrm{answer}\mathrm{if}\mathrm{someone}\mathrm{have}\mathrm{a}\mathrm{idea}\mathrm{he}\mathrm{give}\mathrm{it}...$$

Answered by mathmax by abdo last updated on 12/Aug/20

$$\mathrm{let}\mathrm{take}\mathrm{a}\mathrm{try}\mathrm{with}\mathrm{this}\mathrm{limit}\mathrm{let}{\mathrm{A}}_{\mathrm{n}}=\left(\begin{array}{c}1\frac{\mathrm{x}}{\mathrm{n}}\\ -\frac{\mathrm{x}}{\mathrm{n}}1\end{array}\right)$$$$\mathrm{we}\mathrm{have}{\mathrm{A}}_{\mathrm{n}}=\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}\times \left(\begin{array}{c}\frac{1}{\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}}\frac{\mathrm{x}}{\mathrm{n}\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}}\\ -\frac{\mathrm{x}}{\mathrm{n}\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}}\frac{1}{\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}}\end{array}\right)$$$$\mathrm{let}{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)\left(\begin{array}{c}\frac{1}{{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}\frac{\mathrm{x}}{\mathrm{n}{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}\\ -\frac{\mathrm{x}}{\mathrm{n}{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}\frac{1}{{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}\end{array}\right)$$$$\mathrm{we}\mathrm{have}\det \left(\begin{array}{c}......\\ \end{array}\right)=\frac{1}{{\alpha }_{\mathrm{n}}^2\left(\mathrm{x}\right)}+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2{\alpha }_{\mathrm{n}}^2\left(\mathrm{x}\right)}=\frac{1}{1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}}+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2(1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2})}$$$$=\frac{{\mathrm{n}}^2}{{\mathrm{n}}^2+{\mathrm{x}}^2}+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2+{\mathrm{x}}^2}=1\mathrm{let}\frac{1}{{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}=\cos \alpha \mathrm{and}\frac{\mathrm{x}}{\mathrm{n}{\alpha }_{\mathrm{n}}\left(\mathrm{x}\right)}=\sin \left(\alpha \right)\Rightarrow$$$$\tan \alpha =\frac{\mathrm{x}}{\mathrm{n}}\Rightarrow \alpha =\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\Rightarrow$$$${{\mathrm{A}}_{\mathrm{n}}^{\frac{\mathrm{n}}{\mathrm{x}}}={\alpha }_{\mathrm{n}}\left(\mathrm{x}\right))}^{\frac{\mathrm{n}}{\mathrm{x}}}\times {\left(\begin{array}{c}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha \end{array}\right)}^{\frac{\mathrm{n}}{\mathrm{x}}}$$$$={(1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2})}^{\frac{\mathrm{n}}{2\mathrm{x}}}\times {\left(\begin{array}{c}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha \end{array}\right)}^{\frac{\mathrm{n}}{\mathrm{x}}}\mathrm{if}\mathrm{x}\mathrm{divide}\mathrm{n}\mathrm{we}\mathrm{get}$$$${\mathrm{A}}_{\mathrm{n}}^{\frac{\mathrm{n}}{\mathrm{x}}}={(1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2})}^{\frac{\mathrm{n}}{2\mathrm{x}}}\left(\begin{array}{c}\cos \left(\frac{\mathrm{n}}{\mathrm{x}}\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)\sin \left(\frac{\mathrm{n}}{\mathrm{x}}\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)\\ -\sin \left(\frac{\mathrm{n}}{\mathrm{x}}\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)\cos \left(\frac{\mathrm{n}}{\mathrm{x}}\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)\right)\end{array}\right)$$$${(1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2})}^{\frac{\mathrm{n}}{2\mathrm{x}}}={\mathrm{e}}^{\frac{\mathrm{n}}{2\mathrm{x}}\ln (1+\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2})}\sim {\mathrm{e}}^{\frac{\mathrm{n}}{2\mathrm{x}}\left(\frac{{\mathrm{x}}^2}{{\mathrm{n}}^2}\right)}={\mathrm{e}}^{\frac{\mathrm{x}}{2\mathrm{n}}}\to 1\mathrm{also}\frac{\mathrm{n}}{\mathrm{x}}\arctan \left(\frac{\mathrm{x}}{\mathrm{n}}\right)$$$$\sim \frac{\mathrm{n}}{\mathrm{x}}.\frac{\mathrm{x}}{\mathrm{n}}=1\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}^{\frac{\mathrm{n}}{\mathrm{x}}}=\left(\begin{array}{c}\cos \left(1\right)\sin \left(1\right)\\ -\sin \left(1\right)\cos \left(1\right)\end{array}\right)$$$$\mathrm{and}\mathrm{Tr}\left({\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}^{\frac{\mathrm{n}}{\mathrm{x}}}\right)=2\cos \left(1\right)$$$$\mathrm{if}\mathrm{n}\mathrm{integr}\mathrm{and}\mathrm{x}\mathrm{dont}\mathrm{divide}\mathrm{n}\mathrm{n}=\mathrm{qx}+\mathrm{p}\mathrm{with}0<\mathrm{p}<\mathrm{x}$$$$\mathrm{be}\mathrm{continued}...$$$$$$$$$$

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