System of equations, 3 equations and 3 unknown I have a system of equations that I'm trying to solve, $Mb = x$
$M=\begin{bmatrix}
e^z &e^z  &e^z \\ 
aX_1 &bX_1  &cX_1 \\ 
aX_2 &bX_2  &cX_2 
\end{bmatrix}$
$b = 
\begin{bmatrix}
Q_1 \\ 
Q_2 \\ 
Q_3 
\end{bmatrix}
$
$x = 
\begin{bmatrix}
1 \\ 
X_1 \\ 
X_2 
\end{bmatrix}
$
Expanding this system out to 
$Q_1e^z + Q_2e^z + Q_3e^z = 1$
$aX_1Q_1 + bX_1Q_2 + cX_1Q_3 = X_1$
$aX_2Q_1 + bX_2Q_2 + cX_2Q_3 = X_2$
This reduces further to 
$Q_1 + Q_2 + Q_3 = e^{-z}$
$aQ_1 + bQ_2 + cQ_3 = 1$
$aQ_1 + bQ_2 + cQ_3 = 1$
I am trying to show that $a>c$ but with the 3 equation and 3 unknown system, I am unable to tackle this. 
 A: First, let me note that you have note restrictions on any of your variables, so I will just assume that $z,a,b,c\in\mathbb{R}$.
Second, have you noticed that your second and third equation are the same equation? You can throw one of them away, leaving only two linear equations in three unknowns. But this might be unimportant to your specific question.
I intepret your questions as in that you are trying to prove that $a>c$ for this system of equations, but this is not necessarily true. Here is the general reason why: Suppose that I have some solution $(Q_1,Q_2,Q_3)=(\widetilde{Q_1},\widetilde{Q_2},\widetilde{Q_3})$ to your equations. Then $(Q_1,Q_2,Q_3)=(\widetilde{Q_3},\widetilde{Q_2},\widetilde{Q_1})$ is a solution to the new, but related, system
$$Q_1 + Q_2 + Q_3 = e^{-z},$$
$$\tilde{a}Q_1 + \tilde{b}Q_2 + \tilde{c}Q_3 = 1,$$
where $\tilde{a}=c$, $\tilde{b}=b$, and $\tilde{c}=a$. It follows that if you can show that generally it must be true that $a>c$, then you have automatically proven that $\tilde{a}<\tilde{c}$ in this other system of the exact same form as the original system, which is a contradiction.
To show you something more concrete, suppose for instance that $z=0$, and $b=1$, and say both $a=c=0$. Then your equations are solved by $(Q_1,Q_2,Q_3)=(0,1,0)$, and you have $a\not> c$.
