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If, in $R^4$, $\alpha_1 = (-1,0,1,2)$, $\alpha_2 = (3,4,-2,5)$, $\alpha_1 = (1,4,0,9)$, is given, How can I find a system of homogeneous linear equations with solutions space which is exactly spanned by above three vectors?

I don't know even how to start with it. Can you give me a hint?

smw1991
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  • Try using the definitions of "Span" and "Homogeneous system", do you remember/know these? – Edward Evans Mar 29 '16 at 12:58
  • Yes, I keep thinking that, if I find such an system,(say matrix A), Av = 0 where v is linear combination of three vectors above. However, after that, all I can think is setting unknown quantities and using inner product, which is not on the material in the chapter. – smw1991 Mar 29 '16 at 13:03

1 Answers1

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We want $a,b,c,d$ such that \begin{align} -a+c+2d=0\\ 3a+4b-2c+5d=0\\ a+4b+9c=0 \end{align}

Row-reducing we get \begin{align} a-c-2d=0\\ b+\frac{1}{4}c+\frac{11}{4}d=0 \end{align}

Hence our solution set is $\{(c+2d,-\frac{1}{4}c-\frac{11}{4}d,c,d):c,d\in \mathbb{R}\}$

Thus if we consider the system \begin{align} 4x_1-x_2+4x_3=0\\ 8x_1-11x_2+4x_4=0 \end{align}

we would get a solution spanned by the given vectors.

Bijesh K.S
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