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Question Number 94915 by mathocean1 last updated on 21/May/20

$$\mathrm{We}\mathrm{suppose}\mathrm{in}{\mathbb{R}}^2\mathrm{the}\mathrm{base}(\overrightarrow{i};\overrightarrow{j}).$$$$\mathrm{we}\mathrm{have}\mathrm{these}\mathrm{vectors}:$$$$\overrightarrow{\mathrm{u}}=({\mathrm{m}}^2-\mathrm{m})\overrightarrow{i}+2\mathrm{m}\overrightarrow{j};$$$$\overrightarrow{\mathrm{v}}=(\mathrm{m}-1)\overrightarrow{i}+(\mathrm{m}+1)\overrightarrow{j}\mathrm{m}\in {\mathbb{R}}^{\ast }$$$$1)\mathrm{Determinate}\mathrm{m}\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{system}$$$$(\overrightarrow{\mathrm{u}};\overrightarrow{\mathrm{v}})\mathrm{is}\mathrm{linear}\mathrm{dependant}(\det (\overrightarrow{\mathrm{u}};\overrightarrow{\mathrm{v}})=0)$$$$$$

Answered by mr W last updated on 21/May/20

$$\frac{m^2-m}{m-1}=\frac{2m}{m+1}$$$$\frac{m-1}{m-1}=\frac{2}{m+1}=1$$$$m+1=2$$$$\Rightarrow m=1$$

Answered by mathmax by abdo last updated on 22/May/20

$$\mathrm{u}({\mathrm{m}}^2-\mathrm{m},2\mathrm{m})\mathrm{v}(\mathrm{m}-1,\mathrm{m}+1)$$$$\det (\mathrm{u},\mathrm{v})=0\Rightarrow \left|\begin{array}{c}{\mathrm{m}}^2-\mathrm{m}\mathrm{m}-1\\ 2\mathrm{m}\mathrm{m}+1\end{array}\right|=0\Rightarrow$$$$({\mathrm{m}}^2-\mathrm{m})(\mathrm{m}+1)-2\mathrm{m}(\mathrm{m}-1)=0\Rightarrow$$$$(\mathrm{m}-1)({\mathrm{m}}^2+\mathrm{m}-2\mathrm{m})=0\Rightarrow (\mathrm{m}-1)({\mathrm{m}}^2-\mathrm{m})=0\Rightarrow$$$${(\mathrm{m}-1)}^2\mathrm{m}=0\Rightarrow \mathrm{m}=1\mathrm{or}\mathrm{m}=0.$$$$$$

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