Answer
If the $m\times n$ matrix $\mathbf{A}$ has $n$ pivot columns, then every equation of the form $\mathbf{A}\vec{x}=\vec{b}$ has only basic variables and no free variables. Suppose $\vec{v}_{b}$ solves the equation for some $\vec{b}\neq\vec{0}$. Then by Theorem 6, every other solution for that specific $\vec{b}$ is of the form $\vec{v}_{b}+\vec{v}_{0}$, where $\vec{v}_{0}$ is any solution to the equation $\mathbf{A}\vec{x}=\vec{0}$. But, as stated in the box on page 44, the only solution to $\mathbf{A}\vec{x}=\vec{0}$ is $\vec{x}=\vec{0}$, since the equation has no free variables. Hence, every other equation of the form $\mathbf{A}\vec{x}=\vec{b}$ has at most one solution, given by $\vec{v}_{b}+\vec{v}_{0}=\vec{v}_{b}+\vec{0}=\vec{v}_{b}$.
Work Step by Step
The definitions of basic and free variables are given on page 18. Note that we are proving that, if some equation has a solution, then that solution is unique. However, the only equation we can be certain has any solution at all is $\mathbf{A}\vec{x}=\vec{0}$: it is still possible that no other matrix equation in $\mathbf{A}$ is solvable.
