### 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.

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.

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

0 &-16 &-12 &-44 \\

1 &1 &9 &9 \\}\] You can enter a preliminary result to check whether you're still on the right track.

1 &0 &0 &-2\\

0 &1 &0 &2 \\

0 &0 &1 &1 \\

}#

Row reduction of the matrix can go as follows:

\[

\begin{array}{rcll}

\left(

\begin{array}{cccc}

1 &5 & -4 &4 \\

0 &-16 &-12 &-44 \\

1 &1 &9 &9 \\

\end{array}

\right)

&\sim& \left( \begin{array}{cccc} \color{green}{1} & 5 & -4 & 4 \\ \color{green}{0} & -16 & -12 & -44 \\ \color{green}{0} & -4 & 13 & 5 \\ \end{array} \right) &\color{blue}{\begin{array}{l} \phantom{x} \\ \phantom{y} \\ R_3 - R_1\end{array}}\\ \\

&\sim& \left( \begin{array}{cccc} \color{green}{1} &5 & -4 &4 \\ \color{green}{0} &\color{green}{1} &\frac{3}{4} &\frac{11}{4} \\ \color{green}{0} &-4 &13 &5 \\ \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{31}{4} & -\frac{39}{4} \\ \color{green}{0} & \color{green}{1} & \frac{3}{4} & \frac{11}{4} \\ \color{green}{0} &\color{green}{0} & 16 & 16 \\ \end{array} \right) &\color{blue}{\begin{array}{l} R_1 - 5 R_2\\ \phantom{x} \\ R_3 +4 R_2 \end{array}}\\ \\

&\sim& \left( \begin{array}{cccc} \color{green}{1} & \color{green}{0} & -\frac{31}{4} &-\frac{39}{4}\\ \color{green}{0} & \color{green}{1} & \frac{3}{4} &\frac{11}{4} \\ \color{green}{0} & \color{green}{0} & \color{green}{1} &1 \\ \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} &-2\\ \color{green}{0} &\color{green}{1} &\color{green}{0} &2 \\ \color{green}{0} &\color{green}{0} &\color{green}{1} &1 \\ \end{array} \right) &\color{blue}{\begin{array}{l} R_1+\frac{31}{4} R_3 \\ R_2 -\frac{3}{4} R_3 \\ \phantom{z}\end{array}}

\end{array}\]

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