True or False?
If the homogeneous system $Ax = 0$ has only the trivial solution, then the corresponding non-homogeneous Ax = b is consistent for each b.
My answer is: False.
Since all homogeneous systems are consistent, that $Ax = 0$ has only the trivial solution only says that the Matrix $A$ has the form $m\times n$ where $m\ge n.$
However, it is not sufficient to say if $Ax = b$ is consistent for each $b$, since a counter-example would be:
$$
\begin{eqnarray}
5x &+& 3y &=& 1\\
5x &+& 2y &=& 1\\
5x &+& y &=& 1
\end{eqnarray}
$$
True or False?
Best Answer
Your guess is right, but the explanation is not entirely accurate.
First, note that saying $A x = 0$ has only the trivial solution is actually equivalent to saying that the nullspace of $A$ only contains the null vector or, still equivalently, the columns of $A$ are linearly independent.
Now, the equation $A x = b $ can only have a solution if $b$ is in the column space of $A$. So, as you pointed out, even when the columns of $A$ are linearly independent, still $A x = b $ may have no solution. If $A$ is square, though, then of course $A x = b $ always has a solution when its columns are linearly independent, since then its columns span the whole space.
(This does not apply to your example, though, since $(1,1,1)$ is clearly in the column space of $A$.)