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Show that the system of equations does not have a nonnegative solution: $\left\{\begin{matrix} x + y + z = 0\\ x - y - 2z = 0\\ x + 2y +3z = 0\\ 2x + 2y + z = 1 \end{matrix}\right.$

I want to use Farkas' lemma to this one, so my matrix $A$ and vector $b$ are as follows:

$A = \begin{bmatrix} 1 & 1 &1 \\ 1 & -1 & -2\\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{bmatrix}, \ b = \begin{bmatrix} 0 \\ 0\\ 0 \\ 1 \end{bmatrix}$.

I assume there exists some vector $v = [a, b, c, d] \in \mathbb{R}^4$ so that $v^T A \leq 0$ and $v^Tb > 0$. From the last inequality I get that $d > 0$, and from the first one I obtained a system of inequalities:

$v^TA = \left\{\begin{matrix} a + b + c + 2d \leq 0\\ a - b + 2c + 2d \leq 0\\ a - 2b + 3c + d \leq 0 \end{matrix}\right.$

Is this approach correct? If yes, how do I proceed to find a particular solution?

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    Farkas lemma states, that either your system $Ax=b, x \ge 0$ or $v^Tb > 0, A^Tv \ge 0$. Is that correct? Why you dont solve it directly? – 2017-02-01
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    @user160069 Solve what, the whole system $Ax=b$? Does the existence of one negative solution mean there are no nonnegative solutions? – 2017-02-01
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    If I state that, let's say, $v = [-2, 0, -1, 1]$ satisfies such inequalities, is this enough for the proof? – 2017-02-01
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    Yes, that's enough. – 2017-02-01

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