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Let $U=(x,y,z)\in \mathbb{Z}^3$ and $a(U),b(U),c(U),d(U),e(U),f(U),g,h$ be integers.

Now consider the system of equations: $$x[a(U)]+y[b(U)]+z[c(U)]=g$$ $$x[d(U)]+y[e(U)]+z[f(U)]=h$$

What conditions must this system satisfy so that both equations are equivalent?

My guess is : $$a(U)e(U)-b(U)d(U)=0$$ $$b(U)f(U)-e(U)c(U)=0$$ $$a(U)f(U)-d(U)c(U)=0$$

Am I wrong?

1 Answers1

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Yes you are. If $a=d,b=e,c=f$ but $g\ne h$, your three determinants are all zero but the two equations are not equivalent.

Dario
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  • So we must $g=h$. –  Mar 13 '17 at 22:16
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    Well, in my counterexample, yes. But in general, it can be $g\ne h$. What you are looking for is the concept of linear dependence, which can be defined in $Z$-modules as well as in vector spaces. – Dario Mar 13 '17 at 22:26