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Question Number 183814 by ali009 last updated on 30/Dec/22 | ||
$${determine}\:{eigen}\:{values}\:{and}\:{eigen}\:{vectors}\:{for} \\ $$$${each}\:\lambda\:.\:{and}\:{verify}\:{Ax}=\lambda{x} \\ $$$${A}=\begin{bmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}&{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{bmatrix} \\ $$ | ||
Answered by TheSupreme last updated on 30/Dec/22 | ||
$$\left\{{x}\right\}'={A}\left\{{x}\right\}\:{is}\:{a}\:{rotation}\:{of}\:\frac{\pi}{\mathrm{6}}\:{rads}\:{counterclockwise} \\ $$$${so}\:{there}'{s}\:{no}\:\left\{{X}\right\}\:{such}\:\alpha\left\{{X}\right\}=\left[{A}\right]\left\{{X}\right\} \\ $$ | ||