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I need to find the composition of a function and its inverse so I have the identity function in return. My problem is that I don't seem to undestand how to proceed algebraically.

I have a function with couples, let's say: $f(x,y)=(x+y,\;2x-2y)$
The inverse would be: $f^{-1}(x,y)=(2x-2y,\;x+y)$
Now, to prove that $f^{-1}$ $\circ$ $f$ = identify function is where I'm getting confused.

I've been trying to solve this way: $f^{-1}(f(x,y)) = 2(x+y)-2y,x+(2x-2y)$ because I thought it was the way to proceed but it gives me an incorrect result.

Anyone could point me in the right direction? Thanks!

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    i actually did, i have a pretty poor math background..!2012-11-02

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Your inverse is not correct, and your calculation shows that. To find the inverse, set $(a,b)=f(x,y)=(x+y,2(x-y))$ and solve for $x$ and $y$ in terms of $a$ and $b$. That'll give your your inverse $f^{-1}(a,b)$.

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    @wwwe You should consider accepting the answer, if it really answered your original question. It'll help us both, plus help keep up with site standards.2012-11-23