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I have the system of these three equations:

$$ax = y+z$$ $$by = x+z$$ $$cz = x+y$$

How do I find all $a$, $b$ and $c$ for which the system has real, positive solutions for $x$, $y$ and $z$?

As a comparison, I have a simpler system:

$$ax = y$$ $$by = x$$

For this simpler system, with a substitution I get that $ab = 1$ yields a system that has real positive solutions. I'd like to find something similar for the above system as well. Basically, I'm looking for the relation between $a$, $b$ and $c$, independent from $x$, $y$ and $z$.

This is somewhat similar to this other question, only my equations are different.

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If it's similar to the other question (which indeed it is) you should understand the answer to that question and figure out how it applies to your equations. You can turn this into a homogeneous system and row-reduce the resulting matrix, then figure out what it means for the system to have a nontrivial solution. –  rogerl Jun 12 at 14:58

2 Answers 2

up vote 2 down vote accepted

Multiply the second equation by $c$ to obtain $$bcy=cx+cz=cx+x+y$$ using the third equation, so that $$(bc-1)y=(c+1)x$$

Likewise multiply the first equation by $c$ to obtain equivalently $$(ac-1)x=(c+1)y$$Now multiply this second equation by $c+1$ so that $$(c+1)^2y=(ac-1)(c+1)x=(ac-1)(bc-1)y$$whence $y=0$ or $$c^2+2c+1=abc^2-ac-bc+1$$ so $c=0$ or $$abc-a-b-c=2$$


Note that the condition $c=0$ can be avoided (by symmetric argument) unless $a=b=c=0$


Further note that positive solutions imply $a,b,c\gt 0$ - because e.g. $a=\frac {y+z}x\gt0$.

Also the solution $a=b=c=-1$ has us multiplying by $c+1=0$ which is inadmissible if we want to remain strictly positive.

Suppose we have positive values $a,b,c,x$ - we then find $$y=\frac {ac-1}{c+1}x; z=\frac{ab-1}{b+1}x$$which are positive provided $ac,ab\gt 1$

Writing the condition $abc-a-b-c=2$ in the form $$b(ac-1)=2+a+c$$ the right-hand side is positive, $b$ is positive, so $ac\gt 1$ is enforced by the condition.

So if $a,,b,c\gt 0$ satisfy the condition, there is a unique positive solution for any chosen positive value of $x$.

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how do you deal with the requirement for positive solution ? –  Rene Schipperus Jun 12 at 15:15
    
@ReneSchipperus that's a good question, but as far as I can tell (by trying different values for $a$, $b$ and $c$), I could always get positive solutions. –  Attila O. Jun 13 at 9:19
    
@ReneSchipperus I've added some comments at the end. –  Mark Bennet Jun 13 at 10:12
    
@AttilaO. I hope my extra comments in the answer help you to understand why this is. –  Mark Bennet Jun 13 at 10:13

This is an homogeneous system of the type $\pmb{A}\pmb{x}=\pmb{0}$.

Every homogeneous system has at least one solution, known as the zero solution (or trivial solution), which is obtained by assigning the value of zero to each of the variables. If the system has a non-singular matrix ($\det(\pmb{A}) ≠ 0$) then it is also the only solution. If the system has a singular matrix then there is a solution set with an infinite number of solutions.

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