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

- For $(x_1,y_1)ne(x_2,y_2)$ we should have $F(x_1,y_1)ne F(x_2,y_2)$
- 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

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