The function $f: \mathbb{R}^2 \rightarrow\mathbb{R}^2$ is defined by $f(x,y)=(2x+3y,x+2y)$. Show that $f$ is bijective and find its inverse.
I've got so far:
Bijective = $1-1$ and onto.
$1-1$
if
$(2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$
Then $$2x_1+3y_1=2x_2+3y_2 \qquad (1)$$
$$ x_1+2y_1=x_2+2y_2 \qquad (2)$$
$(1)-(2)$
$$x_1+y_1=x_2+y_2$$
$$ x_1=x_2+y_2-y_1$$
Substituting into equation 1:
$$ 2(x_2+y_2-y_1)+3y_1=2x_2+3y_2$$
$$ y_1=3y_2-2y_2 $$
$$ y_1=y_2$$
Substituting into equation 2:
$$ x_1+2y_1=x_2+2y_1$$
$$ x_1=x_2$$
$$ (2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$$
Thus 1-1
Onto
if $(u,v) \in \mathbb{R}^2$ (codomain) we want $(x,y)$ with $f(x,y)=(u,v)$
$$ (2x+3y,x+2y)=(u,v)$$
$$2x+3y=u$$
$$ x+2y=v$$
Eliminating x:
$$ y=2v-u$$
Substituting:
$$ 2x+3(2v-u)=u$$
$$ 2x+6v-3u=u$$
$$ 2x=4u-6v$$
$$ x=2u-3v$$
Therefore:
$$ f(2u-3v,2v-u)=(u,v)$$
Now how do you find it’s inverse? And is that correct what I have done?
Best Answer
Superhint: If you had started with the second part ("and find its inverse") you could have "coimpletely" avoided the first part calculations.