### Systems of linear equations: System of linear equations and matrices

### Echelon form and reduced row 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 **reduced row 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.

By elementary operations with rows we can always convert a matrix to reduced echelon form. How that will we discuss in the next section work exactly.

-1 &-14 &-2 &26 \\

-1 &-2 &-15 &2 \\}\] You can enter an intermediate result to check whether you're still on the right track.

\matrix{

1 &0 &0 &2\\

0 &1 &0 &-2 \\

0 &0 &1 &0 \\ }\)

Below is a row reduction of the matrix in smallsteps to this result.

\[\begin{aligned} \matrix{1&-2&-2&6\\-1&-14&-2&26\\-1&-2&-15&2\\}&\sim\matrix{1&-2&-2&6\\0&-16&-4&32\\-1&-2&-15&2\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-2&-2&6\\0&-16&-4&32\\0&-4&-17&8\\}&{\blue{\begin{array}{c}\phantom{x}\\\phantom{x}\\R_3+R_1\end{array}}}\\\\ &\sim\matrix{1&-2&-2&6\\0&1&{{1}\over{4}}&-2\\0&-4&-17&8\\}&{\blue{\begin{array}{c}\phantom{x}\\-{{1}\over{16}}R_2\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&0&-{{3}\over{2}}&2\\0&1&{{1}\over{4}}&-2\\0&-4&-17&8\\}&{\blue{\begin{array}{c}R_1+2R_2\\\phantom{x}\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&0&-{{3}\over{2}}&2\\0&1&{{1}\over{4}}&-2\\0&0&-16&0\\}&{\blue{\begin{array}{c}\phantom{x}\\\phantom{x}\\R_3+4R_2\end{array}}}\\\\ &\sim\matrix{1&0&-{{3}\over{2}}&2\\0&1&{{1}\over{4}}&-2\\0&0&1&0\\}&{\blue{\begin{array}{c}\phantom{x}\\\phantom{x}\\-{{1}\over{16}}R_3\end{array}}}\\\\ &\sim\matrix{1&0&0&2\\0&1&{{1}\over{4}}&-2\\0&0&1&0\\}&{\blue{\begin{array}{c}R_1+{{3}\over{2}}R_3\\\phantom{x}\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&0&0&2\\0&1&0&-2\\0&0&1&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2-{{1}\over{4}}R_3\\\phantom{x}\end{array}}} \end{aligned}\]

You can of course also combine steps. The scheme below illustrates this, and the part of the matrix that we have already processed is coloured green.

\[

\begin{array}{rcll}

\left(

\begin{array}{cccc}

1 &-2 & -2 &6 \\

-1 &-14 &-2 &26 \\

-1 &-2 &-15 &2 \\

\end{array}

\right)

&\sim& \left( \begin{array}{cccc} \color{green}{1} & -2 & -2 & 6 \\ \color{green}{0} & -16 & -4 & 32 \\ \color{green}{0} & -4 & -17 & 8 \\ \end{array} \right) &\blue{\begin{array}{l} \phantom{x} \\ R_2 + R_1 \\ R_3 + R_1\end{array}}\\ \\

&\sim& \left( \begin{array}{cccc} \color{green}{1} &-2 & -2 &6 \\ \color{green}{0} &\color{green}{1} &\frac{1}{4} &-2 \\ \color{green}{0} &-4 &-17 &8 \\ \end{array} \right) &\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{3}{2} & 2 \\ \color{green}{0} & \color{green}{1} & \frac{1}{4} & -2 \\ \color{green}{0} &\color{green}{0} & -16 & 0 \\ \end{array} \right) &\blue{\begin{array}{l} R_1 +2 R_2\\ \phantom{x} \\ R_3 +4 R_2 \end{array}}\\ \\

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

\end{array}\]