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Question Number 98250 by bobhans last updated on 12/Jun/20

$$\mathrm{Let}\mathrm{A}=\left(\begin{array}{c}22\\ 13\end{array}\right).\mathrm{Find}\mathrm{a}\mathrm{non}\mathrm{singular}\mathrm{matrix}$$$$\mathrm{P}\mathrm{such}\mathrm{that}{\mathrm{P}}^{-1}\mathrm{AP}\mathrm{is}\mathrm{a}\mathrm{diagonal}\mathrm{matrix}.$$

Commented by john santu last updated on 12/Jun/20

$$\mathrm{find}\mathrm{eigen}\mathrm{vector}\det (\mathrm{A}-\lambda \mathrm{I})=0$$$$\left|\begin{array}{c}2-\lambda 2\\ 13-\lambda \end{array}\right|=0\Rightarrow \lambda =1;4$$$$\mathrm{Eigen}\mathrm{vector}\mathrm{for}\lambda =1$$$$(\mathrm{A}-\mathrm{I})\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)\Rightarrow \left(\begin{array}{c}12\\ 12\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$\mathrm{eigen}-\mathrm{vector}\left[\begin{array}{c}-2\\ 1\end{array}\right]$$$$\mathrm{for}\lambda =4\Rightarrow \left(\begin{array}{c}-22\\ 1-1\end{array}\right)\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$\to \mathrm{x}-\mathrm{y}=0;\mathrm{eigen}-\mathrm{vector}=\left[\begin{array}{c}1\\ 1\end{array}\right]$$$$\therefore \mathrm{P}=\left[\begin{array}{c}-21\\ 11\end{array}\right]\mathrm{such}\mathrm{that}$$$${\mathrm{P}}^{-1}\mathrm{AP}=\left[\begin{array}{c}10\\ 04\end{array}\right]◼$$

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