Given a system of n linear equations. Prove that the system is inconsistent if and only if you can obtain 0 = 1, by using linear combinations.
I do not want to apply theorems from linear algebra here. Instead of it I want to use Farkas' lemma.
Every equation can be rewrited in form of two inequalities with $\leq$ and $\geq$, now I got $2n$ inequalities instead of $n$ equations.
I think that's the moment for me to apply Farkas' lemma, but why I can always get $0 \leq 1$ and $0 \geq 1$?
It is very easy to prove the result from linear algebra (Gauss eliminations). If you prefer to work it out using Farkas lemma then you can do the following: the equation $Ax=b$ fits the best to the first alternative of the Farkas lemma, we only missing the positivity of $x$. To get positivity one can write $x=v-w$, where $v,w\ge 0$. It is the standard rewriting in Linear Programming. You get $$ \begin{pmatrix}A & -A\end{pmatrix}\begin{pmatrix}v\\ w\end{pmatrix}=b,\quad v,w\ge 0. $$ This is exactly the first alternative. If it has no solution then the dual system $$ \begin{pmatrix}A^T\\-A^T\end{pmatrix}y\le\begin{pmatrix}0\\0\end{pmatrix},\quad b^Ty>0 $$ has a solution. The dual system is the same as $$ A^Ty=0,\quad b^Ty>0\qquad\Leftrightarrow\qquad y^TA=0,\quad y^Tb>0. $$ One can always scale $y$ to get $y^Tb=1$.
It means that if you multiply your original equation $Ax=b$ by $y^T$ from the left you will get $$ \underbrace{y^TA}_{=0}x=\underbrace{y^Tb}_{=1}\quad\Rightarrow\quad 0=1. $$