Question

Let $R$ denote the set of real numbers. Let $f:R\times R\to R$ be a bijective function defined by $f(x,y)=(x+y,x-y)$. The inverse function of $f$ is given by

• A. $f^{-1}(x,y)=(\frac{1}{x+y},\frac{1}{x-y})$
• B. $f^{-1}(x,y)=(x-y,x+y)$
• C. $f^{-1}(x,y)=(\frac{x+y}{2},\frac{x-y}{2})$
• D. $f^{-1}(x,y)=\left(2\right(x-y),2(x+y\left)\right)$

Answer 1

The correct option is C $f^{-1}(x,y)=(\frac{x+y}{2},\frac{x-y}{2})$

$f(x,y)=\left(\right(x+y),(x-y\left)\right)$
So let $z_1=x+y$
$z_2=x-y$
$f(x,y)=(z_1,z_2)$
So $f^{-1}(z_1,z_2)=(x,y)$
Adding (i) and (ii)
$z_1+z_2=2x$
$x=\frac{z_1+z_2}{2}$
Subtracting (i) and (ii)
$z_1-z_2=2y$
$y=\frac{z_1-z_2}{2}$
So the inverse function of $F$ is given by
$f^{-1}(z_1,z_2)=(x,y)=(\frac{z_1+z_2}{2},\frac{z_1-z_2}{2})$
Since $z_1$ and $z_2$ are just dummy variables so replacing $z_1$ and $z_2$ by $x$ and $y$ respectively
$f^{-1}(x,y)=(\frac{x+y}{2},\frac{x-y}{2})$

Alternate Method:
Given $f(x,y)=\left(\right(x+y),(x-y\left)\right)$
Take some random ordered pair say $(2,3)$
$f(2,3)=\left(\right(2+3),(2-3\left)\right)=(5,-1)$
Now the correct inverse must map $(5,-1)$ back to $(2,3)$
Trying the options one-by-one we find that only option (c) maps $(5,-1)$ to $(2,3)$.