Differences Between Row Echelon and Reduced Row Echelon
Mia Lopez 
Let $$A =\begin{bmatrix} 1&3&-6&2\\5&12&-24&7\\9&15&-30&6\end{bmatrix}$$
$$RREF(A) =\begin{bmatrix} 1&0&0&-1\\0&1&-2&-1\\0&0&0&0\end{bmatrix}$$ $$REF(A) =\begin{bmatrix} 1&3&-6&2\\0&1&-2&1\\0&0&0&0\\\end{bmatrix}$$
Now, provided this information, it can be found that the basis for the nullspace of $A$ using the $RREF(A)$ is: $$\begin{pmatrix} \begin{bmatrix} 0\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} 1\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$
Using the $REF(A)$ the "basis" for the nullspace is:
$$\begin{pmatrix} \begin{bmatrix} 6\\2\\1\\0 \end{bmatrix}, \begin{bmatrix} -2\\-1\\0\\1 \end{bmatrix}\end{pmatrix}$$
Is the "basis" that I provided for the $REF(A)$ actually a basis? How does the $REF$ and $RREF$ of a matrix differ?
$\endgroup$ 23 Answers
$\begingroup$The main difference is that it is easy to read the null space off the RREF, but it takes more work for the REF.
Applying a row operation to $A$ amounts to left-multiplying $A$ by an elementary matrix $E$. This preserves the null space, as $Av = 0 \iff EA v = 0$ (elementary matrices are invertible). Hence both $A$ and its RREF (and REF) have the same null space, and it is a simple matter to read off the null space from the RREF.
$\endgroup$ $\begingroup$From Williams (source), pg. 348:
The difference between a reduced echelon form and an echelon form is that the elements above and below a leading 1 are zero in a reduced echelon form, while only the elements below the leading 1 need be zero in an echelon form.
Examples and further discussion are given in the above text.
Another great resource is available here.
$\endgroup$ $\begingroup$Row Echelon form(REF) requires backward substitution while Row Reduce Echelon form (RREF) requires no backward substitution.
$\endgroup$
