Show that a linear system of equations has a solution if and only one of the unknowns is set equal to a known value Consider the following linear system
$$
\begin{cases}
Y_{1}+Y_2 & =&2\alpha_1+\gamma_1+\gamma_2\\
Y_{3}+Y_4&=&2\alpha_2+2\gamma_1\\
Y_{5}+Y_6&=&2\alpha_3+\gamma_2+\gamma_3\\
Y_{7}+Y_8&=&2\alpha_4+2\gamma_3\\
Y_1+Y_3+Y_4&=&\alpha_1+2\alpha_2+3\gamma_1\\
Y_2+Y_6&=&\alpha_1+\alpha_3+2\gamma_2\\
Y_5+Y_7+Y_8&=&\alpha_3+2\alpha_4+3\gamma_3\\
\end{cases}
$$
with unknowns $x\equiv (\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3)$.
Question: Could you help me to show that this system has a unique solution with respect to $x$ if and only if one component among $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4, \gamma_1,\gamma_2, \gamma_3\}$ is set equal to a known value?
My attempt: I have tried (without success, could you help if you think this is the right path?) to show that one equation among those listed above is a linear combination of the others. Once that is shown, we are left with a linear system in $6$ equations and $7$ unknowns. How do we go from there?
 A: The way I approached this problem was to write the equations in ${\alpha_i, \gamma_i}$ as a matrix. This is, I think, the best way to determine the linear dependence of your system.
The matrix I got was:
$$
\left[ 
\begin{array}{l}
2& 0& 0& 0& 1& 1& 0\\
0& 2& 0& 0& 2& 0& 0\\
0& 0& 2& 0& 0& 1& 1\\
0& 0& 0& 2& 0& 0& 2\\
1& 2& 0& 0& 0& 3& 0\\
1& 0& 1& 0& 0& 2& 0\\
0& 0& 1& 2& 0& 0& 3
\end{array}
\right]
$$
Then, I just took it's reduced row echelon form, which according to Julia this is: $$
\begin{array}{l}\alpha_1& \alpha_2& \alpha_3& \alpha_4& \gamma_1& \gamma_2& \gamma_3\end{array}
$$
$$
\left[\begin{array}{l}
 1&  0&  0&  0&  0&  0&   1\\
 0&  1&  0&  0&  0&  0&   1\\
 0&  0&  1&  0&  0&  0&   1\\
 0&  0&  0&  1&  0&  0&   1\\
 0&  0&  0&  0&  1&  0&  -1\\
 0&  0&  0&  0&  0&  1&  -1\\
 0&  0&  0&  0&  0&  0&   0
\end{array}
\right]
$$
As can be clearly seen, on the rightmost column of this vector that one parameter must in fact be a free parameter. That is, every other parameter is defined in terms of this free parameter (can actually be any element of the set, but as soon as it's set you're stuck with the choice). Therefore, this system necessarily needs a known value for the value of the free parameter to be uniquely soluble.
