Don't understand what a problem in Artin's "Algebra" is asking for. 
Solve completely the systems $AX =0$ and $AX = B$, where$$A = 
\begin{pmatrix} 
1 & 1 & 0 \\
1 & 0 & 1 \\
1 & -1 & -1 \\
\end{pmatrix}, \quad \text{and} \quad B= 
\begin{pmatrix}
 1 \\ -1 \\ 1\\
\end{pmatrix},$$
  
  
*
  
*in $\mathbb{Q}$;
  
*in $\mathbb{F}_2$;
  
*in $\mathbb{F}_3$;
  
*in $\mathbb{F}_7$.
  

I am confused by the phrasing of the question. Does it mean solve the two systems $AX = 0$ and $AX = B$ simultaneously or separately? What would it even mean to solve them simultaneously here?
More generally, how would I get started on this problem? I've been stuck for quite a long while, and a push from a friendly stranger would be well-appreciated.
 A: Hint


*

*You are being asked to solve the two systems homogeneous $AX=0$ and the non-homogeneous $Ax=b$. Since solving essentially would mean row reduction of $A$ so technically one can solve both systems in one go. By row reducing something like


$$\left[\begin{array}{lrr|l|r} 
1 & 1 & 0 &\color{red}{0}&\color{blue}{1}  \\
1 & 0 & 1 &\color{red}{0}&\color{blue}{-1} \\
1 & -1 & -1 &\color{red}{0}&\color{blue}{1} \\
\end{array}\right]$$
But keep in mind that you may need the solutions for the homogeneous system to discuss solutions for the non-homogeneous system.


*The other part is asking you to find the solution in the respective fields.

A: 
Does it mean solve the two systems $AX = 0$ and $AX = B$ simultaneously or separately?

Separately.

More generally, how would I get started on this problem? 

Have you ever solved linear systems before? At the very worst, you could simply write out $X=\begin{bmatrix}a\\b\\c\end{bmatrix}$ and multiply things out and see what the resulting system looks like.
You could also attempt to compute the inverse matrix of $A$, but beware that the point of the problem is probably that it may not be singular in all the cases given.
Solving the first one will take you toward solving the second one. After all, every solution of $AX=B$ is of the form $k+X$ where $X$ is a single solution to $AX=B$ and $k$ is any solution to $AX=0$.
