# Is a $1 \times n$ matrix already in echelon form?

Take $A = [1, -2, 0, 4]$ and $B=[1, 0, 0, 0]$ for example. I know the conditions to be in row echelon form are:

• The first nonzero element in the row is a $1$ if the row does not consist entirely of zeros (a leading $1$).
• All rows where the elements are zero are grouped to the bottom of the matrix.
• In any two successive rows (say $\text{r}_1$, $\text{r}_2$), the "leading $1$" occurs farther to the right in $\text{r}_2$ than in $\text{r}_1$.
• Each column that has a "leading $1$" has zeros everywhere else in that column.

Based on these conditions, is it safe to say that $A$ and $B$ are in row echelon form or reduced row echelon form?

Any $1\times n$ matrix is automatically in echelon form. However, if the leading entry is not a $1$, it will not be in reduced echelon form, so you'll have to divide through and fix that.