I'm finding myself a bit confused by an example from some lecture notes. They lay out the simple process of finding an inverse of a standard function, say, $y = 2x + 5$, by simply solving for $x$, giving an inverse of $x = \frac{y-5}{2}$, assuming I did the algebra correctly.

One of the examples doesn't seem to make sense, though. It starts with the function $y = 3x - 4z + 2$, and claims that solving for $x$ yields the inverse. This doesn't seem to make much sense, since this is a function from $\mathbb{R^2} \to \mathbb{R}$ (I know this to be the case, and I think the argument is that it isn't one-for-one and this may draw on the rank-nullity theorem, though it'd be great if someone could explain this with more rigor) and there doesn't seem to be a good reason to solve for $x$ instead of $z$.

Is this reasoning correct? I suppose it's possible that there's a typo in the notes and $z$ is simply a constant. But it doesn't seem like this inverse exists.

I'd appreciate any helpful comments, especially on the reasoning above regarding what I believe is an application of rank-nullity. Thanks.

  • $\begingroup$ You can treat $z$ as a parameter. The inverse is about $x$ and $y$. $\endgroup$ – W. mu Jun 15 '18 at 1:31
  • $\begingroup$ Does that imply, then, that we'd fix $z$ and vary $x$ and $y$? It seems to met that we wouldn't be able to allow $z$ to vary if the inverse were only about $x$ and $y$. $\endgroup$ – Matt.P Jun 15 '18 at 1:35
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    $\begingroup$ This is not a single inverse funciton, but a series inverse function with parameter $z$. Similarly, what do you think of the integration with parameter? $\endgroup$ – W. mu Jun 15 '18 at 1:46

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