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I have this linear system:

$\left\{\begin{array}{c} 2x + 3y - 4z = \ 1 \\ 3x - y - 2z = 2 \\ x - 7y - 6z = 0 \end{array}\right.$

I found the following solution:

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \alpha \frac{10}{11}+\frac{7}{11} \\ \alpha \frac{8}{11}-\frac{1}{11} \\ \alpha \end{pmatrix} $

but the correct solution is

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10t+7\\ 8t+5 \\ 11t+7 \end{pmatrix} $

I know that the two solution are equivalent (and correct). Assuming that i don't know the second form, how i can transform the first form into the last form (with only integer coefficents)?

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1 Answer 1

just conveniently look at $z$ and you see $$z=\alpha=11t+7$$ If you substitute this for $\alpha$ in the first form, you get the second form. You may also invert this relationship between $t,\alpha$ to get $$t = \frac{\alpha-7}{11} $$ If you substitute to the second form, you get the first form.

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Of course. But if i don't know the second form how i can linearize the solution? –  Katy23 May 25 '11 at 17:50
    
Apologies, I don't understand this question. What does it mean to "linearize the solution"? –  Luboš Motl May 25 '11 at 17:53
    
I mean that the solution must have only integer coefficients –  Katy23 May 25 '11 at 17:55
    
I see, that was my guess that this is what you wanted. Well, if you had your form only, you would first make the coefficients of $\alpha$ integer by writing $\alpha = 11\beta$ where $11$ was found as the smallest common multiple of the denominators. That would yield $(x,y,z) = (10\beta+7/11,8\beta-1/11,11\beta)$. Then you would shift $\beta$ in such a way that the absolute coefficients are also integer. –  Luboš Motl May 25 '11 at 18:02
    
You want to shift $\beta$ by a multiple of $1/11$, by $k/11$, that makes the absolute coefficients integer as well. $10k+7$ and $8k-1$ have to be a multiple of $11$. $k=7$ just does the job but that's not the only solution: $k=-4$ or any $11n-4$ does the job, too. –  Luboš Motl May 25 '11 at 18:04

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