It's injective if it is a one-to-one function.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 $\displaystyle f^{-1}$

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

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

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.