Can $Ax = b$ have infinitely many solutions for every $b$ in $ℝ^m$? I've been learning about linear independence/dependence and this question was asked: Suppose an m×n matrix A has n pivot columns. Explain why for each b in $ℝ^m$  the equation Ax=b has at most one solution? Hint: Explain why Ax=b cannot have infinitely many solutions.
For reference:
Let A be an m×n matrix.  Then the following statements are logically equivalent:


*

*For each b in $\Bbb R^m$, the equation Ax = b has a solution.

*Each b in $\Bbb R^m$ is a linear combination of the columns of A.

*The columns of A span $\Bbb R^m$.

*A has a pivot position in every row.


I understand that each b in $ℝ^m$ has one solution, because, in the question, every column is a pivot column.  Therefore, there are no free variables, and Ax = b cannot have infinitely many solutions.  However, all this led to my question:
Is there a circumstance where Ax = b has infinitely many solutions for every b in $ℝ^m$, or, if there's a solution for every b in $ℝ^m$, is it always unique (only one)?
Edit: Would this matrix A have infinitely many solutions for every b?
\begin{array}{l}1&0&0&*\\0&1&0&*\\0&0&1&*\end{array}
Asterisk = any number.  If you added vector b to the right hand side of this matrix A, to make it an augmented matrix, there would still be a pivot position in every row, but there would also be a free variable $x_4$.  Would this be a circumstance where Ax = b has infinitely many solutions for every b in $ℝ^m$?

If the $m\times n$ matrix $A$ has rank $n$ ($n$ pivot columns), then $A$ has a left inverse. If $L$ is such a left inverse and $v_1$ and $v_2$ are solutions of $Ax=b$, then $Av_1=b=Av_2$, so
$$
v_1=I_nv_1=LAv_1=LAv_2=I_nv_2=v_2
$$
Hence the system $Ax=b$ has at most one solution.
In another way: the system $Ax=b$ can have infinitely many solutions only if the matrix $A$ has some nonpivot column (in this case the system has either infinitely many solutions or none). Indeed, when you perform Gaussian elimination, pivot columns correspond to unknowns whose value is determined by the vector $b$, whereas nonpivot columns correspond to “free variables” which can be given arbitrary values.
Conversely, if every system $Ax=b$ has a solution, then the matrix $A$ must have $m$ pivot columns, or some system would have no solution. If both conditions hold (that is, $n$ pivot columns and every system has a solution), then $m=n$ and the matrix $A$ is invertible.
