If a homogeneous system has only the trivial solution, is $Ax = b$ consistent?

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?

• What do you mean by "consistent"? Nov 18 '17 at 8:25
• that the linear system has a solution Nov 18 '17 at 11:14
• But the given example has the solution (x = 1/5, y = 0), right? Sep 19 '18 at 10:24
• Possible duplicate of linear algebra homogeneous system Sep 23 '18 at 10:28

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 $A$.)