Finding the solutions of a system with variables 
We have the following system over $\mathbb R$:
$x+y+z=3\\ 
\alpha x+\beta y + \gamma z =1\\
\beta x+\gamma y + \alpha z =1\\
\gamma x+\alpha y + \beta z =1$
For which $\alpha ,\beta, \gamma$ does the above system have a solution ? When will this solution be singular ? (there's no need to find the solutions).

As far as I can see the first equation is irrelevant because it doesn't have any of alpha, beta, gamma. Making a matrix out of the other three equations doesn't really help, the only thing I can make out of it is that $\alpha=1-\beta-\gamma$. I also see that the determinant isn't zero (if $\alpha \neq\beta \neq \gamma $) so it's linearly independent if this is the case, but now I don't know how to continue...
 A: You're almost done.
You already know that $\alpha+\beta+\gamma=1$ -- for example because the sum of the last three equations is $(\alpha+\beta+\gamma)(x+y+z)=3$, which has to be true at the same time as $x+y+z=1$.
Once we set $\alpha=1-\beta-\gamma$, the last equation is a linear combination of the three first ones, so we can forget that. What we have left is the coefficient matrix
$$ \begin{bmatrix} 1 & 1 &1 \\
1-\beta-\gamma & \beta & \gamma \\
\beta & \gamma & 1-\beta-\gamma \end{bmatrix}$$
You also know the determinant of this is nonzero except when $\beta=\gamma=1/3$ -- or so you say; let me just give an argument for that for completeness.
After a few row operations the coefficient matrix becomes
$$ \begin{bmatrix} 1 & 1 &1 \\
1-2\beta-\gamma & 0 & \gamma-\beta \\
\beta-\gamma & 0 & 1-\beta-2\gamma \end{bmatrix}$$
If we set $\beta=p+q$ and $\gamma=p-q$ this is
$$ \begin{bmatrix} 1 & 1 &1 \\
1-3p+q & 0 & -2q \\
2q & 0 & 1-3p-q \end{bmatrix}$$
whose determinant is $-4q^2 - [(1-3p)^2 - q^2] = -[3q^2 + (1-3p)^2]$.
The only way for this determinant to be $0$ is if $q=1-3p=0$ or in other words $\beta=\gamma=1/3$.

All of the above you already know.
In the case where the determinant is $0$ we have $\alpha=\beta=\gamma=1/3$, and so the system of equations have plenty of solutions, such as for example $x=y=z=1$ or $x=3, y=z=0$.
On the other hand, when the determinant is nonzero, the coefficient matrix is invertible, and then we know that the system has a unique solution. (And, by the way, since $x=y=z=1$ is always a solution when $\alpha+\beta+\gamma=1$, that must be it).
So the answer is that the system has at least one solution exactly if $\alpha+\beta+\gamma=1$.
