Let $A \in \mathbb R^{3 \times 3}$ such that the linear system $Ax = (1,1,1)^t$ has a solution but the system $Ax = (1,1,0)^t$ does not.
Which of the following options is correct ?
The system $Ax=(0,0,0)^t$ has a more than $1$ solution;
The system $Ax=(2 2 1)^t$ has no solutions;
For all $0\neq C \in \mathbb R^{3 \times 3}$ the system $ACx=(1,1,1)^t$ has a solution
If $D$ is row equivalent to $A$, the system Dc=91 1 1) has a solution
$r(A)=2$
What I can understand is that $A$ is not $0$ so $r(A)!=0$. Also because $Ax=(1 1 0)$ has no solution $r(A)<3$, so the first option is not true.
But I'm not sure about the other options