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given these vectors : $v_1=(1,-2,0,3,1 ),v_2=(2,-3,2,5,-3),v_3=(1,-2,1,2,-2)$ over the field $\mathbb{Q}$

What is the homogeneous system of equations such that the $span(v_1,v_2,v_3)$ is equal to the set of all the solutions to the system.

I know that $x_1 = 1,x_2 =-2, x_3 =0 ,x_4 =3 ,x_5=1$ is a solution and so on, but how to get the system of equation ?!!

I am studying linear algebra and get stuck on this question.

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If $w \in \text{Span}(v_1,v_2,v_3)$, then $\exists x,y,z \in \Bbb Q$ such that $$w = xv_1+yv_2+zv_3.$$ So in terms of equations you want the following system to be consistent: \begin{align*} x+2y+z & =w_1\\ -2x-3y-2z & =w_2\\ 2y+z & =w_3\\ 3x+5y+2z & =w_4\\ x-3y-2z & =w_5. \end{align*} So find the condition for consistency and you will get your system.

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These vectors are linearly independent. Find vectors $v_4$ and $v_5$ of $\mathbb{Q}^5$ such that $\{v_1,v_2,v_3,v_4,v_5\}$ is a basis. Express each vector of $\mathbb{Q}^5$ as $\alpha v_1+\beta v_2+\gamma v_3+\delta v_4+\varepsilon v_5$. The the system that you're after is$$\left\{\begin{array}{l}\delta=0\\\varepsilon=0.\end{array}\right.$$Express this condition with respect to the standard basis, and you're done.

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