My approach.
Let $a$ denote a leading entry and $b$ be any value.
The possible echelon forms of a $3\times 3$ matrix are:
$$\begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & a\end{bmatrix}, \begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} a & b & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & a & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.$$
If these are correct, we can pretty much augment any matrix of the appropriate size to the right of these matrices and these matrices would remain in echelon form right? Even if suppose for the fifth matrix I augment it with something that will make the system inconsistent, i.e.
$$\begin{bmatrix} 0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} 0 \\ 5 \\ 0\end{bmatrix}$$
is still in echelon form. Or am I mistaken somewhere?
Best Answer
here are the rrefs of $3 \times 3:$
of rank $3:$ $\quad \pmatrix{1&0&0\\0&1&0\\0&0&1}$
of rank $2: \quad\pmatrix{1&0&x\\0&1&x\\0&0&0} , \pmatrix{1&x&0\\0&0&1\\0&0&0}, \pmatrix{0&1&0\\0&0&1\\0&0&0}$
of rank $1: \quad\pmatrix{1&x&x\\0&0&0\\0&0&0}, \pmatrix{0&1&x\\0&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\0&0&0}$
of rank $0: \quad \pmatrix{0&0&0\\0&0&0\\0&0&0}$
where $x$ represents an arbitrary number. hopefully this is complete.