Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.