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$f:R^{2}\rightarrow R^{2}$ where $f(x,y)=(2x+y,x+y)$.

So I know that the function is one to one and onto and don't need help proving that. But with its inverse, I am a little confused. My professor said the inverse is $g(x,y)=(x-y,2x-y)$, and I want to know how he knew that. He doesn't really explain how got the inverse, just states what the inverse is. So, how do you know what the inverse is? Is there some way to solve for it?

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You have $f(x,y)=(2x+y,x+y)$. You can view the inverse as looking for $(x',y')=(2x+y,x+y)$ and viewing $x,y$ as functions of $x',y'$ which gives two simultaneous equations in two unknowns: $$x'=2x+y \\ y'=x+y$$

Now solve these for $x,y$ and get $$x=x'-y' \\ y=2y'-x'$$

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Thats legit, thanks a lot! – USC Feb 18 '13 at 4:34
I did it the same way you did, it should just be y=2y'-x instead of the other way. But the method works. – USC Feb 18 '13 at 4:34
To claim "this is incorrect" without point out the mistake is incorrect...imo, at least. – DonAntonio Feb 18 '13 at 4:35
@USC: You are right. Thanks and good on you for checking. I have fixed. – Ross Millikan Feb 18 '13 at 4:38

You can write the mapping in matrix form as $$f(x,\ y)=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$$ The inverse of a $2\times 2$ matrix is easy to compute. In this case, the inverse matrix is given by $$\begin{pmatrix}1 & -1 \\ -1 & 2\end{pmatrix}$$ which corresponds to the function $f^{-1}(x,\ y)=(x-y,\ -x+2y)$.

By the way, your inverse is off by a bit in the second coordinate.

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