An infeasible Phase I problem and its Farkas certificate

The third right-hand side is negative, so $D=\mathrm{diag}(1,1,-1)$ gives $\tilde{A}=DA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 1& -1\end{array}\right]$ and $\tilde{b}=Db=(1,1,3{)}^{\mathsf{T}}$. The Phase I point $x=(1,1,0)$, $u=(0,0,1)$ satisfies $\tilde{A}x+u=\tilde{b}$ and has objective $1$.

The Phase I dual vector $y=(-1,-1,1)$ satisfies ${\tilde{A}}^{\mathsf{T}}y=(0,0,-1{)}^{\mathsf{T}}\le 0$ and ${\tilde{b}}^{\mathsf{T}}y=1$. For any Phase I feasible $(x,u)$, its objective ${\mathbf{1}}^{\mathsf{T}}u$ is at least ${\tilde{b}}^{\mathsf{T}}y=1$ by multiplying the constraints by $y$ and using ${\tilde{A}}^{\mathsf{T}}y\le 0$, $u\ge 0$, and $y\le \mathbf{1}$. Thus the displayed feasible point attains the Phase I optimum $1>0$.

The original Farkas vector is $q=-D^{\mathsf{T}}y=(1,1,1)$. Direct calculation gives $A^{\mathsf{T}}q=(0,0,1{)}^{\mathsf{T}}\ge 0$ and $b^{\mathsf{T}}q=1+1-3=-1<0$. Hence no $x\ge 0$ can satisfy $Ax=b$, as also follows from [F2].