Proper way to "discuss a system" I am doing exercises to increase my math skills and I found this one:
Let $A$ be a real parameter. Discuss the system:
$$x - y + 3z = 3A - 1$$
$$2x - (A+1)y -Az = -3$$
I did consider the three cases where $A=0$, $A=1$ and $A$ is different from both $0$ and $1$, and I got three different solutions. And I am not at all asking for those as I am pretty confident about them but my problem is that when I asked for the correction, I was told that it involved using matrix, determinants,... but could not access it.
Did I answer to "discuss the system" in a proper way or did I miss/forget something ?
Are there alternative ways to discuss a system?
 A: You can do without matrices. Solve the first equation with respect to $x$:
$$x=y-3z+3A-1$$
and substitute in the second:
$$(1-A)y - (A+6)z = -1 - 6A$$
What can we say?
If $A=1$, then we get a "fixed" value for $z$, namely $z=1$, so the solutions are the triples $(h-1,h,1)$, where $h$ is arbitrary: in fact the first equation becomes simply $x=y-1$.
If $A\ne1$, you can solve the equation with respect to $y$:
$$y=\frac{(A+6)z-1-6A}{1-A}$$
and you can plug it in the first equation, obtaining solutions where the value of $z$ is arbitrary (it's just annoying to write it down).
A: Well for this system you have two equations in three variables so it is an underdetermined system i.e. the system yields either no solutions or infinitely many solutions. 
@egreg is right, while a matrix is not necessary to analyze the system, if we arrange the system in an appropriate matrix and row reduce accordingly we can glean a few insights: 
$$\left( \begin{matrix} 1 & -1 & 3 & -1+3A \\ 2 & -1-A & -A & -3 \end{matrix} \right) \implies \left( \begin{matrix} 1 & 0 & \frac{A + 6}{A - 1} + 3 & \frac{6A - 1}{A -1} + 3A -1 \\ 0 & 1 & \frac{A+6}{A-1} & \frac{6A + 1}{A - 1} \end{matrix} \right) $$
The above was obtained through Wolfram|Alpha here.
Namely that: 
1) For all $A \in \mathbb{R}$ there are infinitely many solutions for this system 
2) If $A = 1$ the matrix reduces to $\left( \begin{matrix} 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & 1\end{matrix} \right)$ or $z = 1$ and $ x- y = - 1$. This would yield probably the most stable form of the system. Where $x$ is solved in terms of $y$ and $z$ is fixed.
3) If $A = 0$ the matrix reduces to $\left( \begin{matrix} 1 & 0 & -3 & -2 \\ 0 & 1 & -6 & -1\end{matrix} \right)$ or $x - 3z = -2 $ and $y - 6z = -1$.
