Exercise 4.1.6

Multiply the equations by $y_1,y_2,y_3=1,1,-1$. Now the equations add to $0=1$ so there is no solution. In subspace language, $y=(1,1,-1)$ is in the left nullspace. $Ax=b$ would need $0=\left(y^{T}A\right)x=y^{T}b=1$ but here $y^{T}b=1$.