Is there only a unique way to decompose a matrix as a series of multiplication of elementary matrix ? (RREF and invertible matrix) We know that for each given matrix, there exists only one unique matrix in row reduced echelon form.
We also know that the transformations of a matrix to another can be decomposed to a series of elementary matrices:
$$A=E_nE_{n-1}E_{n-2}\dots E_1 (P) \text{ (where P is the original matrix)}$$
As a result, I would like to ask:

*

*Does it exist only one way to transform a matrix to the RREF ?

*Is the matrix resulted from $E_nE_{n-1}E_{n-2}\dots E_1$ that transforms a matrix to its RREF unique?

We also know that to find the inverse of an invertible matrix, we can also apply a series of multiplication of elementary matrices.
In that case, is $E_nE_{n-1}E_{n-2}\dots E_1$ unique in finding the inverse matrix?
 A: *

*The sequence of transformations is not unique. For example you can insert an operation, then undo it by applying the inverse of the operation.


*The matrix $E_nE_{n-1}\cdots E_1$ is not unique unless the original matrix $P$ is of full rank. For example, consider $P:=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, whose RREF is $A:=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Try writing $A$ as the product $EP$, for $E:= \begin{pmatrix}a&b\\c&d\end{pmatrix}$. After expanding $EP$, you will find that the only constraints are $a+b=1$ and $c+d=0$. This means you have some freedom in selecting $a, b, c, d$.


*If $P$ has full rank, then the RREF $A$ is the identity matrix. So if $E$ satisfies the equation $A=EP$, then $E$ must be the inverse of $P$ and is therefore unique.
A: Try row reducing $\begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}$ in two ways: first, by just adding row 1 to row 2, and second, by swapping the rows, adding (the new) row 1 to (the new) row 2, and then scaling row 1 by -1. Pretty sure you get different products of $E$'s.
In more detail, consider the matrix $\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ (implementing the first RREF) and the matrix product $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (implementing the second). Are they the same?
As a metamathematical note, I came up with this example from a general intuition that the product of $E$s may depend on which rows you choose to use when clearing out entries from a given column.  Often you have some choice in that, and apparently it does make a difference.
