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I would like to ask whether the following matrix is a row echelon matrix (and reduced row-echelon matrix).
The below are $1 \times 1$ matrices.
(a) $\begin{pmatrix} 0\end{pmatrix}$
(b) $\begin{pmatrix} 1\end{pmatrix}$
(c) $\begin{pmatrix} 2\end{pmatrix}$

The below are $n\times1$ matrices, where $n\in \mathbb{R}$ and $n \geq 2$
(d) $\begin{pmatrix} 0 & 0 & \cdots & 0\end{pmatrix}$

Here are my opinions. I would like for my opinions be confirmed (or to be countered if mine is wrong):
(a) It is a row echelon form and reduced row echelon form.
(b) It is a row echelon form and reduced row echelon form. The entry $1$ is the leading one.
(c) It is a row echelon form, but it is not a reduced row echelon form, since there is no leading $1$ before $2$
(d) It is a row echelon form and reduced echelon form. \

(For (d), it is inspired by " Is a $1 \times n$ matrix already in echelon form? ".)

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1 Answer 1

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According to Linear Algebra and Its Applications by Lay, p. 13

A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties:

  1. All nonzero rows are above any rows of all zeros.
  2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.
  3. All entries in a column below a leading entry are zeros.

If a matrix in echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form):

  1. The leading entry in each nonzero row is 1.
  2. Each leading 1 is the only nonzero entry in its column.

A leading entry of a row refers to the leftmost nonzero entry (in a nonzero row).

So, yes you are correct based on the above.

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