Prove that $F(x,y)=(x-4y,2x+3y)$ is bijective

Question

I saw the solution of this similar exercise Prove that $F(x,y) =(2x+y, x+4y)$ is bijective, But we didn't study ker Yet, I want an elementary solution, please

I need to prove that :$F(x,y)=(x-4y,2x+3y)$ is injective and surjective.

Answer 1

For proving bijectiveness you have to show two things

1. For $(x_1,y_1)\ne(x_2,y_2)$ we should have $F(x_1,y_1)\ne F(x_2,y_2)$
2. For any $(a,b)$, there exists $(x,y)$, such that $F(x,y)=(a,b)$

For proving 1 we show that If $F(x_1,y_1)=F(x_2,y_2)$, then we must have $(x_1,y_1)=(x_2,y_2)$. Now \begin{equation} x_1-4y_1=x_2-4y_2\tag{1} \end{equation}

\begin{equation} 2x_1+3y_1=2x_2+3y_2\tag{2} \end{equation} Then $(2)-(1)\times 2$ yields $11y_1=11y_2$, which implies $y_1=y_2$, which again implies that $x_1=x_2$.

For proving 2 we show that the system of equations

\begin{equation} x-4y=a\tag{3} \end{equation}

\begin{equation} 2x+3y=b\tag{4} \end{equation}

has a solution for any $a,b\in\Bbb{R}$. $(4)-(3)\times2$ yields $11y=b-2a\implies y=\frac{b-2a}{11}\implies x=\frac{4b+3a}{11}$. Thus there indeed is a solution.

Hence $F(x,y)$ is both one-to-one and onto, hence bijective.

Answer 2

Put $(u,v)=(x-4y,2x+3y)$. Then $u=x-4y$ and $v=2x+3y$. This gives $x=u+4y$ and so $v=2(u+4y)+3y = 2u+7y$ or $y=\frac{1}{7}(v-2u)$.

Swap $u,v$ with $x,y$.

Now you have it $f^{-1}(x,y) = (x+4y,\frac{1}{7}(y-2x))$.