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Although I am currently studying fractional linear transformations in complex analysis, I suddenly had the need to find the "solution" of a system of linear equations but I could not. Unless I have made some mistake in the algebra on the way, the system is as follows:

$ac' = a'c$

$ad \ ' +bc' = a'd + b'c$

$bd \ ' = b'd$

Observe from the first and last equation that we have equal ratios. So if I can show that $ab' = a'b$ then this should imply that the variables are multiples of each other by some fixed constant. I am suspecting that I am to use the second equation in some manner (hopefully not brute force using any of the other two) to obtain the desired equality.

Any help will be greatly appreciated!

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So $a,a',b,b'c,c',d$ and $d'$ are the variables you are trying to solve for? Then these equations are not linear. They are homogeneously quadratic, since every term has degree two in unknowns quantities.. – alex.jordan Dec 8 '12 at 19:21
up vote 1 down vote accepted

You can eliminate to of the variables and be left with a single equation relating the other 6. But there will still be infinitely many solutions. To eliminate $a'$, I'd multiply the second equation by $c$ so that you can swap out all instances of $a'c$. Then I'd do the same for $b'$, multiplying by $d$ and swapping out the $b'd$.

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Is it possible to show what I want to show? I don't want to solve this, I just want to show that $ab' = a'b$ – user44069 Dec 8 '12 at 19:30
No. It's possible for $a=b'=0$ and $a'=b=1$. Just try entering those and see what would be required of the remaining variables. – alex.jordan Dec 8 '12 at 19:38

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