# Is it okay to determine pivot positions in a matrix in echelon form, not in reduced echelon form?

Definition

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):
4. The leading entry in each nonzero row is 1.
5. Each leading 1 is the only nonzero entry in its column.

Definition A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.

Source: Linear Algebra and Its Applications, David C. Lay

In the solution, $\begin{bmatrix} 1 &4 &5 &-9 &7 \\ 0 &2 &4 &-6 &-6 \\ 0 &0 &0 &-5 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix} \space \text{are not in reduced echelon form}\\$
The author haven't yet checked if the positions in 2 and -5 satisfy the definition of pivot position when the echelon matrix is reduced to a reduced echelon form.
I think if it becomes the reduced echelon form, the columns above the pivot positions, the author determined, can be nonzeros and the entries in pivot positions, author dermined, can be $0$ and or have nonzero entry to its left.
So author's determining from the above matrix that leading 1, 2, -5 are pivot positions without reducing the echelon is wrong.

Is there something I'm missing? Or doesn't it have a chance of getting a different pivot position when the echelon form is changed into a reduced echelon form?

In a $m×n$ matrix in echelon form of a linear system for some positive integers m, n, let the leading entries $(■)$ have any nonzero value, and the starred entries $(☆)$ have any value including zero.

Leading entries $■$s in $R_1$ and $R_2$ in an echelon matrix can become leading 1 in a reduced echelon matrix through dividing them by $■$, and the entry ☆ in $R_1$ above $■$ in $R_2$ can be $0$ by subtracting a multiple of $■$.

So $R_1$ and $R_2$ in a matrix in echelon form becomes as follows:
$\begin{array}{rcl} R_1\space & [■ ☆\cdots ☆☆☆☆]\\ R_2\space & [0 ■\cdots ☆☆☆☆]\end{array} \qquad ~ \begin{array}{rcl} R_1\space & [1 0\cdots ☆☆☆☆]\\R_2 &[0 1\cdots ☆☆☆☆] \end{array}$

For all integers k with $2≤k<m$, $R_k$, $R_{k+1}$ in the echelon matrix can be expressed as
$R_{k}\space$ $[0 \cdots 0 ■☆☆\cdots ☆]$
$R_{k+1}$ $[0 \cdots 0 0 ■☆\cdots ☆]$.

Subtracting a multiple of leading entry of $R_{k+1}$ from $R_k$ can make the entry above leading $■$ in $R_{k+1}$ be zero, and the leading $■$s in $R_k$, $R_{k+1}$ can be 1 through dividing the rows by leading entry $■$s.

So the rows in echelon matrix become the following in reduced $m×n$ echelon matrix:
$\begin{array}{rcl} R_{k}\space & [0 \cdots 0 ■☆☆\cdots ☆]\\ R_{k+1} & [0 \cdots 0 0 ■☆\cdots ☆]\\ \end{array} \qquad \begin{array}{rcl} R_{k} & [0 \cdots 0 1 0 ☆\cdots ☆]\\ R_{k+1} & [0 \cdots 0 0 1 ☆\cdots ☆]\\ \end{array}$

Hence, it's found that leading 1s in reduced echelon form of $m×n$ matrix of a linear system correspond to the locations of the leading non-zero values in a $m×n$ matrix in echelon form of the linear system.

When the matrix is in echelon form, for each non-zero row $R_i$, you can divide the row by its leading non-zero value. that makes the leading value $1$. Next for each row $R_n$ above $R_i$, you can subtract $R_i$ multiplied by the entry in row $R_n$ in the same column as that leading $1$ from $R_n$. This results in $R_n$ having $0$ in that column. If you follow this procedure starting with the first row and going down, by the time you are done, the matrix will be reduced echelon form, and guess what! Those leading $1$s that define the pivot points are exactly the locations of the leading non-zero values in each row back before it was reduced.
$$\begin{bmatrix} \color{red}1 &4 &5 &-9 &7 \\ 0 &\color{red}2 &4 &-6 &-6 \\ 0 &0 &0 &\color{red}{-5} &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ First divide each row by its leading non-zero value $$\begin{bmatrix} \color{red}1 &4 &5 &-9 &7 \\ 0 &\color{red}1 &2 &-3 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ Subtract $4$ times Row 2 from Row 1: $$\begin{bmatrix} \color{red}1 &0 &-3 &3 &19 \\ 0 &\color{red}1 &2 &-3 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ Subtract $3$ times row 3 from row 1, and add 3 times row 3 to row 2: $$\begin{bmatrix} \color{red}1 &0 &-3 &0 &19 \\ 0 &\color{red}1 &2 &0 &-3 \\ 0 &0 &0 &\color{red}1 &0 \\ 0 &0 &0 &0 &0 \\ \end{bmatrix}$$ And now, we are in reduced echelon form. See that the the pivot points from the definition are in the same locations as the echelon from leading non-zero values.