I more or less understand what I'm trying to do here, I'm just doubtful of my linear system manipulation; I feel like I'm going wrong somewhere. Can someone let me know if I'm doing something wrong?
We have $f$: $\mathbb R$$^3$ $\to$ $\mathbb R$$^3$ defined by
$f(x,y,z) = (x - y + 2z, -x +y -z, 2x - y + z)$
I want to show that $\forall$$y$ $\in$ $\mathbb R$$^3$, $\exists$$x$ $\in$ $\mathbb R$$^3$ such that $f(x) = y$
So,
take any $(x,y,z)$$\in$ $\mathbb R$$^3$, we have $(a,b,c)$$\in$ $\mathbb R$$^3$ such that $f(x,y,z) = (a,b,c)$
$\Rightarrow$ $(x - y + 2z, -x +y -z, 2x - y + z) = (a,b,c)$
$\Rightarrow$
1. $x - y + 2z = a \\
2. -x +y -z = b \\
3. 2x - y + z = c$
(do 1<-2, 3<-2)):
$\Rightarrow$ 1) $z = a + b\\ 2) -x +y -z = b \\ 3) x = c + b$
(replace the values of x and z into 2):
$\Rightarrow$ 1) $z = a + b\\ 2) y = b + x + z = b + (c + b) + (a + b) = a+ 3b + c \\ 3) x = c + b$
Thus, $\forall$ $(x,y,z)$, we have a $(a,b,c)$ for which $f(x,y,z) = (a,b,c)$
I'm almost 100% I've done something wrong, but I just don't know how to go about doing this the 'proper way'; what is there to fix?
Quick follow up question: $\forall$ $(a,b,c)$$\in$$\mathbb R$$^3$, what are the conditions of $(a,b,c)$ such that $(a,b,c)$$\in$ f($\mathbb R$$^3$).
Wouldn't the conditions be the same as the ones found when looking for surjectivity?
Any help is greatly appreciated. Apologies for the poor formatting, I couldn't find a way of making nice long curly brackets. Thank you.
Your method is correct, you directly solved the system and prove that always we can obtain a solution for all $(a,b,c)\in\mathbb{R^3}$ and thus that $f$ is surjective.
As an alternative we can show that the system
$$\begin{bmatrix}1&-1&2\\-1&1&-1\\2&-1&1 \end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$$
has always solution, that is matrix $A$ is not singular, indeed $\det(A)=-1\neq 0$.