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Question Number 153227 by Eric002 last updated on 05/Sep/21
![let D= [((v 5)),(((1/3) m)) ] find number (v) and (m) such that D^2 =5I (I=identity matrix)](Q153227.png)
$${let}\:{D}=\begin{bmatrix}{{v}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:{m}}\end{bmatrix}\:{find}\:{number}\:\left({v}\right)\:{and} \\ $$$$\left({m}\right)\:{such}\:{that}\:{D}^{\mathrm{2}} =\mathrm{5}{I}\:\:\:\:\:\left({I}={identity}\:{matrix}\right) \\ $$
Answered by puissant last updated on 05/Sep/21
![D^2 = [((v^2 +(5/3) 5v+5m)),(((v/3)+(m/3) (5/3)+m^2 )) ]= [((5 0)),((0 5)) ] ⇒ { ((v^2 =5−(5/3))),((v+m=0)) :}, { ((v+m=0)),((m^2 =5−(5/3))) :} ⇒ m=−v v^2 =((10)/3) ⇒ v=(√((10)/3)) ; m=−(√((10)/3)).. or v=−(√(((10)/3) )) ; m=(√((10)/3))..](Q153228.png)
$${D}^{\mathrm{2}} =\begin{bmatrix}{{v}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\mathrm{5}{v}+\mathrm{5}{m}}\\{\frac{{v}}{\mathrm{3}}+\frac{{m}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{3}}+{m}^{\mathrm{2}} }\end{bmatrix}=\:\begin{bmatrix}{\mathrm{5}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{5}}\end{bmatrix} \\ $$$$\Rightarrow\:\begin{cases}{{v}^{\mathrm{2}} =\mathrm{5}−\frac{\mathrm{5}}{\mathrm{3}}}\\{{v}+{m}=\mathrm{0}}\end{cases},\:\begin{cases}{{v}+{m}=\mathrm{0}}\\{{m}^{\mathrm{2}} =\mathrm{5}−\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\:{m}=−{v}\: \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{3}}\:\Rightarrow\:{v}=\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}\:\:;\:\:{m}=−\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}.. \\ $$$${or}\:{v}=−\sqrt{\frac{\mathrm{10}}{\mathrm{3}}\:}\:;\:\:{m}=\sqrt{\frac{\mathrm{10}}{\mathrm{3}}}.. \\ $$
Commented by Eric002 last updated on 05/Sep/21
$${well}\:{done} \\ $$$$ \\ $$
Answered by King1 last updated on 06/Sep/21
$$ \\ $$