How is the following matrix in reduce echelon form?

Question

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.

Answer 1

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.

Answer 2

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.