Help with anti-image matrix First of all, I am very sorry but I don't know the mathematics terminology in English, so I'll try to explain as good as i can but i will probably do some mistakes since it's not my native language.
I have this problem: given the endomorphism $f_k(x,y,z) = (x+y+kz, x+ky, 2x+(k+1)y+kz)$ from $\mathbb{R}^3$ to $\mathbb{R}^3$, determine for $k \in \mathbb{R}$


*

*a base of $\mathop{\rm Im} f_k$ and a base of $\mathop{\rm Ker} f_k$,

*the set $f^{-1}(0,2,2)$.


I have a good idea on how to solve the first one, but I am blocked on the second one. I'd like some help if possible.
Again, sorry for the many mistakes, and thanks a lot for your patience and attention.
 A: Definition: $f^{-1}(0,2,2) = \{ (x,y,z): f(x,y,z) = (0,2,2) \}$. So, you have a system of equations:
\begin{align*}
0 &= x + y + kz, \\
2 &= x + ky, \\
2 &= 2x + (k+1)y + kz.
\end{align*}
Note that the sum of the first two equations gives you the second, so the system is not regular and will have many solutions or none at all.
I'll assume this is enough of a hint. If you need help with solving the system, feel free to ask in the comments.
Edit (solving the system)
I got $x = 2-ky$ from the second equation, and then I substituted $x$ in the first one, obtaining
$$y = \frac{2+kz}{k-1}.$$
Then I've put that in $x = 2-ky$ and got
$$x = -\frac{2+k^2z}{k-1}.$$
Wolfram|Alpha gives the same solution, so it should be correct.
Obviously, for $k = 1$ this solution is not acceptable, as we have a division by zero. This means that we have to solve it separately for $k = 1$ (we could ignore the third equation, as it is still the sum of the first two, but I'll write it anyway):
\begin{align*}
0 &= x + y + z, \\
2 &= x + y, \\
2 &= 2x + 2y + z.
\end{align*}
Now, we have $x = 2 - y$. Putting that into the first equation, we get
$$z = -2.$$
There are no more equations to use, so we're done here.
So, our final solution is:
\begin{align*}
f^{-1}(0,2,2) &= \left\{ \left(-\frac{2+k^2z}{k-1}, \frac{2+kz}{k-1}, z \right): z \in \mathbb{R} \right\}, \quad \text{for $k \ne 1$}, \\
f^{-1}(0,2,2) &= \left\{ \left(2 - y, y, -2 \right): y \in \mathbb{R} \right\}, \quad \text{for $k = 1$}.
\end{align*}
We could have solved by $(x,z)$ (see here) or by $(y,z)$ (see here), which would give us different parametrizations of the same solution.
To check that these solutions are really correct, try calculating $f(x,y,z)$ of the above parametrized solutions (both cases $k \ne 1$ and $k = 1$). As a result, you should get $(0,2,2)$ in all cases (for all $k$ and $z$).
