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Show-that-E-x-y-z-R-3-x-2y-z-0-is-a-subspace-vector-of-which-we-will-determine-one-base-please-help-sirs-




Question Number 79111 by mathocean1 last updated on 22/Jan/20
Show that  E={(x,y,z) ∈ R^3   /  x−2y+z=0}  is a subspace vector of which we  will determine one base.  please help sirs...
$$\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{E}=\left\{\left({x},\mathrm{y},{z}\right)\:\in\:\mathbb{R}^{\mathrm{3}} \:\:/\:\:{x}−\mathrm{2}{y}+{z}=\mathrm{0}\right\} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{subspace}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{which}\:\mathrm{we} \\ $$$$\mathrm{will}\:\mathrm{determine}\:\mathrm{one}\:\mathrm{base}. \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{sirs}… \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
x−2y +z =0 ⇒x=2y−z ⇒(x,y,z)=(2y−z,y,z)  =(2y,y,o) +(−z,0,z) =y(2,1,0) +z(−1,0,1) =y u^→  +zv^→  ⇒  E is a vectorial plane with base B=(u^→ ,v^→ )  u^→ (2,1,0) and v^→ (−1,0,1)
$${x}−\mathrm{2}{y}\:+{z}\:=\mathrm{0}\:\Rightarrow{x}=\mathrm{2}{y}−{z}\:\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{2}{y}−{z},{y},{z}\right) \\ $$$$=\left(\mathrm{2}{y},{y},{o}\right)\:+\left(−{z},\mathrm{0},{z}\right)\:={y}\left(\mathrm{2},\mathrm{1},\mathrm{0}\right)\:+{z}\left(−\mathrm{1},\mathrm{0},\mathrm{1}\right)\:={y}\:\overset{\rightarrow} {{u}}\:+{z}\overset{\rightarrow} {{v}}\:\Rightarrow \\ $$$${E}\:{is}\:{a}\:{vectorial}\:{plane}\:{with}\:{base}\:{B}=\left(\overset{\rightarrow} {{u}},\overset{\rightarrow} {{v}}\right) \\ $$$$\overset{\rightarrow} {{u}}\left(\mathrm{2},\mathrm{1},\mathrm{0}\right)\:{and}\:\overset{\rightarrow} {{v}}\left(−\mathrm{1},\mathrm{0},\mathrm{1}\right) \\ $$
Commented by mathocean1 last updated on 22/Jan/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Commented by mathmax by abdo last updated on 23/Jan/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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