Question Number 153227 by Eric002 last updated on 05/Sep/21
$${let}\:{D}=\begin{bmatrix}{{v}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:{m}}\end{bmatrix}\:{find}\:{number}\:\left({v}\right)\:{and} \\ $$$$\left({m}\right)\:{such}\:{that}\:{D}^{\mathrm{2}} =\mathrm{5}{I}\:\:\:\:\:\left({I}={identity}\:{matrix}\right) \\ $$
Answered by puissant last updated on 05/Sep/21
$${D}^{\mathrm{2}} =\begin{bmatrix}{{v}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\mathrm{5}{v}+\mathrm{5}{m}}\\{\frac{{v}}{\mathrm{3}}+\frac{{m}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{3}}+{m}^{\mathrm{2}} }\end{bmatrix}=\:\begin{bmatrix}{\mathrm{5}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{5}}\end{bmatrix} \\ $$$$\Rightarrow\:\begin{cases}{{v}^{\mathrm{2}} =\mathrm{5}−\frac{\mathrm{5}}{\mathrm{3}}}\\{{v}+{m}=\mathrm{0}}\end{cases},\:\begin{cases}{{v}+{m}=\mathrm{0}}\\{{m}^{\mathrm{2}} =\mathrm{5}−\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\:{m}=−{v}\: \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{3}}\:\Rightarrow\:{v}=\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}\:\:;\:\:{m}=−\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}.. \\ $$$${or}\:{v}=−\sqrt{\frac{\mathrm{10}}{\mathrm{3}}\:}\:;\:\:{m}=\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}.. \\ $$
Commented by Eric002 last updated on 05/Sep/21
$${well}\:{done} \\ $$$$ \\ $$
Answered by King1 last updated on 06/Sep/21
$$ \\ $$