# Determining solvability of system of equations

I have a bizarre system of 4 equations I'm trying to solve. I'm pretty sure they're unsolvable but I'm not sure how to prove it mathematically. I've tried using Matlab's "Solve" function and received back some empty symbols but I'm not sure if that's definitive proof that the system is unsolvable.

Here are the equations:

$$5.25 = a/b$$

$$9.0781= a/b + a/d$$

$$10.1875 = a/b+a/c$$

$$11.1875 = a/b+a/c+a/d$$

Is this system unsolvable? How can you prove unsolvability in a case like this?

• Welcome to MSE. Why do you say that this system is linear? – José Carlos Santos Feb 16 '20 at 8:28
• Back-substitute for $a/b$, $a/c$ and $a/d$ in turn and see what you end up with. – amd Feb 16 '20 at 8:35
• Well, I ended up with 11.1875 = 14.0156... so I guess It's not solvable. – Payton Grenich Feb 16 '20 at 8:54

Let us divide all equations by $$a$$ and group terms in a same side of each equation ; the resulting system can be given the following expression :
$$\left[\begin{array}{cccc}-\alpha&1&0&0\\ -\beta&1&0&1\\ -\gamma&1&1&0\\ -\delta&1&1&1\end{array}\right]\left[\begin{array}{c}1/a\\1/b\\1/c\\1/d\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]$$
(where $$\alpha=5.25$$, $$\beta=9.0781$$, etc.)
As the determinant of the square matrix is $$-\alpha+\beta+\gamma-\delta=2.8281 \neq 0$$, the only solution to such a system would be the null vector. But this is impossible to have for example $$1/a=0$$ with a real number $$a$$.