How to find inverse of function $f(x, y)$?

Question

I am aware of the method to find inverse function $f^{-1}(x)$ of $f(x)$, which is

• Replace $f(x)$ with $y$
• Switch $x$'s and $y$'s
• Solve for $y$
• Replace $y$ with $f^{-1}(x)$

the above method worked well until I found following problem,

Let $f\colon \mathbb R\rightarrow \mathbb R$ denotes a bijective function defined as;
$f(x, y) = (x + y + 1, x - y - 1)$,
then, find the inverse $f^{-1}(x, y)$

can anybody please tell me how to solve such kind of problems?

Answer 1

• Replace $f\left(x,y\right)$ by $\left(u,v\right)$ resulting in: $x+y+1=u$ and $x-y-1=v$

• Switch $x$ and $u$ and switch $y$ and $v$ resulting in: $u+v+1=x$ and $u-v-1=y$

• Solve for $u$ and $v$ resulting in: $u=\frac{1}{2}x+\frac{1}{2}y$ and $v=\frac{1}{2}x-\frac{1}{2}y-1$

• Replace $\left(u,v\right)$ with $f^{-1}\left(x,y\right)$ resulting in: $f^{-1}\left(x,y\right)$=$\left(\frac{1}{2}x+\frac{1}{2}y,\frac{1}{2}x-\frac{1}{2}y-1\right)$

So, same procedure. This gives you the inverse of function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ defined by $f\left(x,y\right)=\left(x+y+1,x-y-1\right)$. I think (as Git Gud) that is what you are after.

Answer 2

$$f(x, y) =(x + y + 1,x - y - 1)$$ $$x\to x+y+1,y\to x-y-1$$ $$x+y+1=a,x-y-1=b\Rightarrow x=\frac{a+b}{2}, y=\frac{a-b-2}{2}$$ $$f^{-1}(x, y) = \left(\frac{x + y}{2},\frac{x-y-2}{2}\right)$$