Let F be a mapping from $R^3$ to $R^3$ be defined by $F(x_1,x_2,x_3) =(x_1-x_2, x_2-x_3, x_3-x_1)$
A) What are the conditions in order that (a,b,c)$\in Imf$ ?
B) What are the conditions in order that (a,b,c)$\in Ker f$.
Clearly I got for B) (a,b,c)$\in kerf$ when a=b=c.
I m stuck for A).
I want to find a, b ,C satisfying
$x_1 -x_2 =a$, $x_2-x_3=b$ $x_3-x_1=c$.
The matrix of $F$ is $$ A=\begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} $$ and Gaussian elimination gives \begin{align} \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{bmatrix} && R_2\gets R_2+R_1 \\ &\to \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} && R_3\gets R_3+R_2 \\ \end{align} From this we infer that a basis of the image is formed by the first two columns of $A$ and that a basis of $\ker F$ consist of the vector $(1,1,1)$. So a vector $(a,b,c)$ belongs to $\ker F$ if and only if $a=b=c$.
For a vector $(a,b,c)$ to belong to $\operatorname{im}F$ we need the linear system represented by the matrix $$ \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & b \\ 0 & -1 & c \end{bmatrix} $$ has a solution. Again an elimination gives \begin{align} \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & b \\ 0 & -1 & c \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & a \\ 0 & 1 & a+b \\ 0 & -1 & c \end{bmatrix} &&R_2\gets R_2+R_1 \\ &\to \begin{bmatrix} 1 & 0 & a \\ 0 & 1 & a+b \\ 0 & 0 & a+b+c \end{bmatrix} &&R_3\gets R_3+R_2 \\ \end{align} so $a+b+c=0$.
As you see, you don't need to guess. The two eliminations could have been performed together: \begin{align} \left[\begin{array}{ccc|c} 1 & 0 & -1 & a \\ -1 & 1 & 0 & b \\ 0 & -1 & 1 & c \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 0 & -1 & a \\ 0 & 1 & -1 & a+b \\ 0 & -1 & 1 & c \end{array}\right] && R_2\gets R_2+R_1 \\ &\to \left[\begin{array}{ccc|c} 1 & 0 & -1 & a \\ 0 & 1 & -1 & a+b \\ 0 & 0 & 0 & a+b+c \end{array}\right] && R_3\gets R_3+R_2 \\ \end{align} giving the same information.
Hint: What can you say about $a + b + c$ if $(a,b,c) \in {\rm Im \ } f$?
Another hint: you may like to use the fact that the dimension of the image plus the dimension of the kernel equals the dimension of the domain, i.e. three.
Hint. You can firstly find basis of $Ker(f)$. Then complete this basis to be basis in $\mathbb{R}^3$. And then find image of these completing vectors. They will be a basis of $Im(f)$.