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

Mathematics Asked by user835019 on February 17, 2021

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.

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 $$begin{equation} x_1-4y_1=x_2-4y_2tag{1} end{equation}$$

$$begin{equation} 2x_1+3y_1=2x_2+3y_2tag{2} end{equation}$$ 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

$$begin{equation} x-4y=atag{3} end{equation}$$

$$begin{equation} 2x+3y=btag{4} end{equation}$$

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

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

Correct answer by QED on February 17, 2021

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 $$f^{-1}(x,y) = (x+4y,frac{1}{7}(y-2x))$$.

Answered by Wuestenfux on February 17, 2021