# Proving the existence of a solution of a matrix (Linear Algebra)

Prove the following statement or give a counterexample if it is false.

Let $A$ be an $m$ by $n$ matrix. if the reduced row echelon form of $A$ has a free column, then there must be some vector $b$ such that $Ax=b$ has no solution.

What i tried

I let A be the following matrix \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}

This matrix seems to satisfy all the condition required for a matrix $A$ and we also know that it will always have a solution no matter what vector $B$ is because it does not have an all zero row.Hence it is a counterexample to disprove the above statement. Is my solution correct. Could anyone explain Thanks

If the reduced row-echelon form of $A$ has a free row (a row with no pivot), then there is some vector $b$ such that $Ax = b$ has no solution.
If the reduced row-echelon form of $A$ does not have a free row, then there is no vector $b$ such that $Ax = b$ has no solution. That is, the linear map $T(x) = Ax$ is onto.