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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)?

1 Answers 1

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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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    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.2011-05-25