Suppose we are given a $4\times3$ matrix $A$ and a column $b$ such that $Ax=b$ always has a unique solution. What is the reduced row echelon form of $A$?
I think that since the solution is unique, there must be pivot positions in every column. This forces the 4th row of A to be zero. So the answer appears to be
\begin{pmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{pmatrix}
Is this right?
But what if $b$ has $1$ as the last coordinate? Then the system would be inconsistent. We are given that $Ax=b$ does have a solution, but this still raises this question: is the following statement true?
For any $4\times3$ matrix $A$ and any column $b$ whose last coordinate is nonzero, the system $Ax=b$ is inconsistent.
If this statement is true, then the column $b$ we are given cannot have nonzero last coordinate.
For the first par, you are right. If $Ax= b$ has a unique solution, then the reduce echelon form of $A$ will be $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ This will give the reduce echelon form of the augmented matrix $(A|b)$ to be $$\left(\begin{array}{ccc|c}1&0&0&\lambda_1\\0&1&0&\lambda_2\\0&0&1&\lambda_3\\0&0&0&0\end{array}\right)$$ where $\lambda_i$ is your solution. If the last line of the augmented matrix is not entirely $0$, then the system is inconsistent. But it need not be true for the initial problem.
The following statement is false.
Consider $$A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&1\end{pmatrix}\text{ and } b=\begin{pmatrix}\lambda_1\\\lambda_2\\\lambda_3\\\lambda_3\end{pmatrix}$$ The last column of $b$ is non zero, but the system is consistent.