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Question Number 148222 by BHOOPENDRA last updated on 26/Jul/21

	Answered by iloveisrael last updated on 26/Jul/21
	$v_3 =λv_1 +αv_2 (((−3)),(( 4)),(( 7)) ) = ((( λ)),(( 0)),((−λ)) ) + ((( 0)),((2α)),((2α)) ) { ((λ=−3)),((α=2)) :}⇒v_3 =−3v_1 +2v_2$
	$v_{3} = λ v_{1} + α v_{2}$ $(\begin{matrix} - 3 \\ 4 \\ 7 \end{matrix}) = (\begin{matrix} λ \\ 0 \\ - λ \end{matrix}) + (\begin{matrix} 0 \\ 2 α \\ 2 α \end{matrix})$ ${\begin{cases} λ = - 3 \\ α = 2 \end{cases} \Rightarrow v_{3} = - {3 v}_{1} + {2 v}_{2}$
	Commented by BHOOPENDRA last updated on 26/Jul/21

	$w h a t a b o u t t h e o t h e r t w o s i r ?$
	Answered by Olaf_Thorendsen last updated on 26/Jul/21
	$(ii) Let v = ((x),(y),(z) ) ∈ W then ∃(α,β,γ)∈R^3 \ v = αv_1 +αv_2 +γv_3 v = αv_1 +βv_2 +γ(−3v_1 +2v_2 ) v = (α−3γ)v_1 +(β+2γ)v_2 Let a = α−3γ and b = β+2γ then ∃(a,b)∈R^2 \ v = av_1 +bv_2 That proves that (v_1 ,v_2 ) generates W. (iii) We solve the equation λv_1 +μv_2 = 0_W (1) ⇔ λ ((1),(0),((−1)) )+μ ((0),(2),(2) ) = ((0),(0),(0) ) ⇔ (λ,μ) = (0,0) We canno′t find (λ,μ)∈R^2 −{(0,0)} such that (1) is true. That proves that v_1 and v_2 are linearly independent.$
	$(ii)$ $Let v = (\begin{matrix} x \\ y \\ z \end{matrix}) \in W$ $then \exists (α, β, γ) \in R^{3} ∖ v = α v_{1} + α v_{2} + γ v_{3}$ $v = α v_{1} + β v_{2} + γ (- 3 v_{1} + 2 v_{2})$ $v = (α - 3 γ) v_{1} + (β + 2 γ) v_{2}$ $Let a = α - 3 γ and b = β + 2 γ$ $then \exists (a, b) \in R^{2} ∖ v = {a v}_{1} + {b v}_{2}$ $That proves that (v_{1}, v_{2}) generates W .$ $(iii)$ $We solve the equation$ $λ v_{1} + μ v_{2} = 0_{W} (1)$ $\Leftrightarrow λ (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}) + μ (\begin{matrix} 0 \\ 2 \\ 2 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$ $\Leftrightarrow (λ, μ) = (0, 0)$ $We {canno}^{'} t find (λ, μ) \in R^{2} - {(0, 0)}$ $such that (1) is true .$ $That proves that v_{1} and v_{2} are$ $linearly independent .$
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