Answer
See answer below
Work Step by Step
From part (c) in the Invertible Matrix Theorem we have $Ax=0$ has only the trivial solution so $REF(A)$ contains a pivot in every column.
The linear system can be solved by back-substitution for every $b \in R^n$.
Assume that there are two solutions $x_1$ and $x_2$ to the system $Ax=b$
$$Ax_1=b$$
$$Ax_2=b$$
$$\rightarrow A(x_1-x_2)=0$$
$$x_1-x_2=0$$
$$x_1=x_2$$
Hence, there is only one solution to the linear system $Ax=b$
