# Thread: Injective function and it's inverse.

1. ## Injective function and it's inverse.

Define $\displaystyle f:\Re^2 \rightarrow \Re^2$ by $\displaystyle f(x,y)=(3x+2y,-x+5y)$. Show that f is injective, and find it's inverse $\displaystyle f^{-1}$
It's injective if it is a one-to-one function.

$\displaystyle f(x_1,y_1)=(3x_1+2y_1,-x_1+5y_1)$
$\displaystyle f(x_2,y_2)=(3x_2+2y_2,-x_2+5y_2)$

I need to show that $\displaystyle x_1=x_2$, $\displaystyle y_1=y_2$ if $\displaystyle f(x_1,y_2)=f(x_2,y_2).$

$\displaystyle f(x_1,y_2)=f(x_2,y_2)$

$\displaystyle (3x_2+2y_2,-x_2+5y_2)=(3x_1+2y_1,-x_1+5y_1)$

Comparing the two sides gives $\displaystyle x_1=x_2$ and $\displaystyle y_1=y_2$

Hence f is injective. I think this proof is right.

The problem arises when I try and find the inverse:

$\displaystyle x=3x+2y$
$\displaystyle y=-x$

and:

$\displaystyle -x+5y=y$
$\displaystyle -x=-4y$
$\displaystyle y=\frac{x}{4}$

I don't think this accomplishes anything but this clearly isn't the right method.

2. Yop,

I personnally think you easily get confused if you use the same names for the new variables.

$\displaystyle f(x,y)=(3x+2y,-x+5y) \Leftrightarrow (x,y)=f^{-1}(3x+2y,-x+5y)=f^{-1}(u,v)$

so let $\displaystyle u=3x+2y$ and $\displaystyle v=-x+5y$

solve for x and y with respect to u and v. this would give something like :
$\displaystyle \left\{\begin{array}{ll} x=\frac{5u}{17}-\frac{2v}{17} \\ y=\frac{u}{17}+\frac{3v}{17} \end{array} \right.$
(check the calculations, I did it quite quickly and it's not the most important part)

hence :
$\displaystyle (x,y)=f^{-1}(u,v)=\left(\frac{5u}{17}-\frac{2v}{17} ~,~ \frac{u}{17}+\frac{3v}{17}\right)$

hey, this is the inverse function !

3. Hey thanks Moo! I solved the simultaneous equations myself so I know where it's all coming from and got what you did!!!

I have been confused once before by a sudden change of variables with the same name. I'll look out for it in the future!