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

We suppose in R^2 the base (i^→ ;j^→ ). we have these vectors: u^→ =(m^2 −m)i^→ +2mj^→ ; v^→ =(m−1)i^→ +(m+1)j^→ m ∈ R^∗ 1)Determinate m for which the system (u^→ ;v^→ ) is linear dependant( det(u^→ ;v^→ )=0)

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

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

((m^2 −m)/(m−1))=((2m)/(m+1)) ((m−1)/(m−1))=(2/(m+1))=1 m+1=2 ⇒m=1

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

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

u(m^2 −m,2m) v(m−1,m+1) det(u,v)=0 ⇒ determinant (((m^2 −m m−1)),((2m m+1)))=0 ⇒ (m^2 −m)(m+1)−2m(m−1) =0 ⇒ (m−1)(m^2 +m −2m) =0 ⇒(m−1)(m^2 −m) =0 ⇒ (m−1)^2 m =0 ⇒m=1 or m=0 .

$$\mathrm{u}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m},\mathrm{2m}\right)\:\:\mathrm{v}\left(\mathrm{m}−\mathrm{1},\mathrm{m}+\mathrm{1}\right) \\ $$$$\mathrm{det}\left(\mathrm{u},\mathrm{v}\right)=\mathrm{0}\:\Rightarrow\begin{vmatrix}{\mathrm{m}^{\mathrm{2}} −\mathrm{m}\:\:\:\:\:\:\:\:\:\mathrm{m}−\mathrm{1}}\\{\mathrm{2m}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{m}+\mathrm{1}}\end{vmatrix}=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m}\right)\left(\mathrm{m}+\mathrm{1}\right)−\mathrm{2m}\left(\mathrm{m}−\mathrm{1}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{m}^{\mathrm{2}} +\mathrm{m}\:−\mathrm{2m}\right)\:=\mathrm{0}\:\Rightarrow\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{m}^{\mathrm{2}} −\mathrm{m}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{m}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{m}\:=\mathrm{0}\:\Rightarrow\mathrm{m}=\mathrm{1}\:\mathrm{or}\:\mathrm{m}=\mathrm{0}\:\:. \\ $$$$ \\ $$

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