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Question Number 136885 by BHOOPENDRA last updated on 27/Mar/21

Answered by bramlexs22 last updated on 27/Mar/21

$${\lambda }^3-\left(\mathrm{trace}\mathrm{A}\right){\lambda }^2+\left(\begin{array}{c}\mathrm{minor}\mathrm{of}\mathrm{the}\mathrm{terms}\\ \mathrm{on}\mathrm{the}\mathrm{leading}\mathrm{diag}\mathrm{A}\end{array}\right)\lambda -\det \left(\mathrm{A}\right)=0$$$${\lambda }^3-3{\lambda }^2+(\left|\begin{array}{c}12\\ 21\end{array}\right|+\left|\begin{array}{c}10\\ 11\end{array}\right|+\left|\begin{array}{c}12\\ -11\end{array}\right|)\lambda -3=0$$$${\lambda }^3-3{\lambda }^2+\lambda -3=0$$$$\mathrm{by}\mathrm{Caley}-\mathrm{Hamilton}\mathrm{teorem}$$$${\mathrm{A}}^3-{3\mathrm{A}}^2+\mathrm{A}-3\mathrm{I}=0$$$$\mathrm{multiply}\mathrm{by}{\mathrm{A}}^{-1}$$$${\mathrm{A}}^2-3\mathrm{A}+\mathrm{I}-{3\mathrm{A}}^{-1}=0$$$${3\mathrm{A}}^{-1}={\mathrm{A}}^2-3\mathrm{A}+\mathrm{I}$$$${3\mathrm{A}}^{-1}=\mathrm{A}(\mathrm{A}-3\mathrm{I})+\mathrm{I}$$$${3\mathrm{A}}^{-1}=\left(\begin{array}{c}-4-24\\ 30-2\\ -302\end{array}\right)+\left(\begin{array}{c}100\\ 010\\ 001\end{array}\right)$$$${3\mathrm{A}}^{-1}=\left(\begin{array}{c}-3-24\\ 31-2\\ -303\end{array}\right)$$$${\mathrm{A}}^{-1}=\frac{1}{3}\left(\begin{array}{c}-3-24\\ 31-2\\ -303\end{array}\right)$$

Commented by BHOOPENDRA last updated on 27/Mar/21

$$thankssir$$

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