How is the following matrix in reduce echelon form?

Consider the matrix

$$\left( \begin{matrix} 1 & 4 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix} \right)$$ According to my textbook, this matrix is in reduced echelon form. However the rules of reduce echelon form state that all other elements in a column that contains a leading 1 are zero.

Would the 4 at position (1,2) not break that rule and make this matrix not be in reduced echelon form?

Thank you.

• ??? That $4$ is not a "leading $1$". Look up the definition of "leading $1$"... Sep 25, 2018 at 14:53
• Yup, just mixed it up Sep 25, 2018 at 14:58

Well, what's the problem then? In every column that contains a leading $$1$$ all other elements are zeros. The $$4$$ is in the second column which has no leading $$1$$. Don't get confused between rows and columns.

• Thank you, I mixed up the row and column, I see how it works now! Sep 25, 2018 at 14:57

Let us define "pivotal column" as the column containing the pivot.

Now the column having element 4 is NOT a pivotal column.

According to echelon form, only pivotal column has all elements (in the column) zero, except the pivot itself.

Since second column is not pivotal, it is perfectly a row-reduced echelon form.

• Thank you! I see how it's reduced now, that makes a lot of sense! Sep 25, 2018 at 14:58