Echelon form and reduced echelon form

Systems of linear equations and matrices: Systems and matrices

Echelon form and reduced echelon form

In the examples of elimination of variables in augmented matrices which were discussed before, the following concepts play such an important role that we give them with a name.

Echelon form By the leading element of a row in a matrix, we mean the first element of the row (from the left) that is distinct from zero.

A matrix is in echelon form if it has the following two properties:

all of the elements in the rows below a leading element, in the column in which this leading element resides as well as in the columns to the left of it, are zero;
null rows, also called zero rows (rows all of whose elements are zero), only appear below all other rows.

In other wordsA matrix has a staircase form: the null rows are at the bottom, and the leading element of each row is strictly on the right side of the leading element of the rows above it.

Upper triangular formWe say that a matrix is in upper triangular form (or just upper triangular) if all #(i,j)#-entries with #i\gt j# are equal to zero. Matrices in echelon form are always in upper triangular form. The entries of the matrix down under that are zero thus form a staircase.

Example

What are the guiding elements below #5\times 3# matrix?

\[\matrix{1 & 4 & 2 \\ 0 & 2 & 6 \\ 0 & 0 & 7 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \\ }\]

The leading element of row 1 is at position 1.

The leading element of row 2 is at position 2.

The leading element of row 3 is at position 3.

The leading element of row 4 is at position 1.

The leading element of row 5 is at position 3.

New example

The leading elements of a matrix in echelon form correspond to unknowns of the associated system of equations that can be written as a linear combination of unknowns with a higher index. If we can apply row reduction to bring a matrix in echelon form, we will have made a step forward towards the solution of a system of linear equations.

As we will see later on, we can use elementary row operations to transform a matrix into echelon form. But we can go further, until the reduced echelon form defined below.

Reduced echelon form A matrix is in row reduced echelon form, or reduced echelon form, if it has the following three properties:

the matrix is in echelon form;
its leading elements are all equal to #1#;
all the elements above a leading element are equal to zero.

A reduced echelon form is recognized by the following properties:

In each row the first element from the left that is not equal to #0# (the leading element), equals #1#; all other elements of the column containing this leading element are zero.
Each row that does not consist of zeros only (a non-null row) starts with more zeros than the previous row. In particular all null rows (rows with only zeros) appear at the bottom.

The matrix \[\left(\begin{array}{ccccc}1 & -1 & -2\\ 0 & 1 & 3\\\end{array}\right)\] is in echelon form, but not in reduced echelon form: the matrix satisfies the first two requirements of the reduced echelon form (the matrix is in echelon form, and all the leading elements are equal to #1#), but not the last condition "all elements above a leading element are equal to zero". After all, in the second column above the leading element #1# of the second row the number #-1#, which is not equal to zero, appears.

By elementary operations with rows we can always convert a matrix to reduced echelon form. How that works will be discussed later.

Determine the reduced row echelon form for the matrix \[\matrix{1 &3 & 4 &-5 \\
-1 &13 &-8 &-27 \\
-1 &1 &11 &-3 \\}\] You can enter a preliminary result to check whether you're still on the right track.

#\matrix{
1 &0 &0 &1\\
0 &1 &0 &-2 \\
0 &0 &1 &0 \\
}#

Row reduction of the matrix can go as follows:
\[
\begin{array}{rcll}
\left(
\begin{array}{cccc}
1 &3 & 4 &-5 \\
-1 &13 &-8 &-27 \\
-1 &1 &11 &-3 \\
\end{array}
\right)
&\sim& \left( \begin{array}{cccc} \color{green}{1} & 3 & 4 & -5 \\ \color{green}{0} & 16 & -4 & -32 \\ \color{green}{0} & 4 & 15 & -8 \\ \end{array} \right) &\color{blue}{\begin{array}{l} \phantom{x} \\ R_2 + R_1 \\ R_3 + R_1\end{array}}\\ \\
&\sim& \left( \begin{array}{cccc} \color{green}{1} &3 & 4 &-5 \\ \color{green}{0} &\color{green}{1} &-\frac{1}{4} &-2 \\ \color{green}{0} &4 &15 &-8 \\ \end{array} \right) &\color{blue}{\begin{array}{l} \phantom{x}\\ \frac{1}{16} R_2\\ \phantom{z}\end{array}}\\ \\
&\sim& \left( \begin{array}{cccc} \color{green}{1} & \color{green}{0} & \frac{19}{4} & 1 \\ \color{green}{0} & \color{green}{1} & -\frac{1}{4} & -2 \\ \color{green}{0} &\color{green}{0} & 16 & 0 \\ \end{array} \right) &\color{blue}{\begin{array}{l} R_1 - 3 R_2\\ \phantom{x} \\ R_3 - 4 R_2 \end{array}}\\ \\
&\sim& \left( \begin{array}{cccc} \color{green}{1} & \color{green}{0} & \frac{19}{4} &1\\ \color{green}{0} & \color{green}{1} & -\frac{1}{4} &-2 \\ \color{green}{0} & \color{green}{0} & \color{green}{1} &0 \\ \end{array} \right) &\color{blue}{\begin{array}{l} \phantom{x}\\ \phantom{y}\\ \frac{1}{16} R_3\end{array}}\\ \\
&\sim& \left( \begin{array}{cccc} \color{green}{1} &\color{green}{0} &\color{green}{0} &1\\ \color{green}{0} &\color{green}{1} &\color{green}{0} &-2 \\ \color{green}{0} &\color{green}{0} &\color{green}{1} &0 \\ \end{array} \right) &\color{blue}{\begin{array}{l} R_1- \frac{19}{4} R_3 \\ R_2 +\frac{1}{4} R_3 \\ \phantom{z}\end{array}}
\end{array}\]

New example