Injective function and it's inverse.

**Showcase_22** · November 19th 2008, 09:29 AM

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

It's injective if it is a one-to-one function.

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

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

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

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

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

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

The problem arises when I try and find the inverse:

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

and:

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

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

**Moo** · November 19th 2008, 10:34 AM

Yop,

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

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

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

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

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

hey, this is the inverse function !

**Showcase_22** · November 19th 2008, 11:48 AM

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!

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