How to prove/disprove - If an inhomogeneous system of equations has more equations than variables, it has no solution.

Question

If an inhomogeneous system of equations has more equations than variables, it has no solution.

Proof: Let $I$ be an inhomogeneous system of equations:

\begin{align} a_{11}x_1&=b_1\\ a_{12}x_1&=b_2 \quad \text{ with }a_{11},a_{12}=1 \end{align} Case 1: $b_1=b_2$
Both equations are equivalent.

Case 2: $b_1 \neq b_2$

In order to solve this system of equations, the equation $x_1=b_1=b_2=x_1$ must hold. This is obviously not possible, because $b_1\neq b_2$. Therefore, the system of equations has no solution. $_\blacksquare$

Is this a proper proof? It's seems a bit too simple.

Answer 1

The system

$$x+y=2$$ $$2x+2y=4$$ $$3x+3y=6$$ has $(1/1)$ as a solution although we have three equations but only two variables. So, the claim is false.

Answer 2

Yes this is a proper proof but not of your original statement. To be frank it is simple because the system you started with is simple. In the general case your statement is wrong or at least highly missleading. If you consider $Ax=b$ where $A$ is a matrix, the existence of at least one solution depends on the rank of $A$ being equal to the rank of the augmented matrix $A|b$.