# Row Echelon Form of a Matrix

If we have a matrix $A$ that represents a system of linear equations, we can reduce $A$ to what is known as **Row Echelon Form** (often times abbreviated as REF) in order to solve the system. We must first look at the definition for a matrix to be in this form.

Definition: Let $A$ be an $m \times n$ matrix. Then $A$ is said to be in Reduced Echelon Form if $A$ satisfies the following three properties.1. All of the rows that do not consist entirely of zeroes will have their first nonzero entries be $1$ which we defined as leading $1$s.2. For any two rows that are not entirely comprised of zeroes, the leading $1$ in the row below occurs farther to the right than the leading $1$ in the higher rows.3. Any rows consisting entirely of zeroes are placed at the bottom of the matrix. |

For example, the following matrices are in REF (as you should verify):

(1)Given a matrix $A$, we can use the elementary row operations to convert any matrix to REF.

## Example 1

**Use elementary row operations to put matrix $A = \begin{bmatrix} 2 & 4 & 6 \\ 2 & 4 & 2\\ 1 & 3 & 1 \end{bmatrix}$ in REF.**

Our first step operation is to take row $1$ and multiply it by $\frac{1}{2}$ ($\frac{1}{2}R_1 \to R_1$):

(2)Now let's multiply row $2$ by $\frac{1}{2}$ ($\frac{1}{2}R_2 \to R_2$):

(3)Now let's take row $2$ and subtract row $1$ from it ($R_2 - R_1 \to R_2$):

(4)Now let's interchange rows $2$ and $3$ ($R_2 \leftrightarrow R_3$):

(5)Now let's take row $2$ and subtract row $1$ from it ($R_2 - R_1 \to R_2$):

(6)Lastly let's multiply row $3$ by $-\frac{1}{2}$ ($-\frac{1}{2}R_3 \to R_3$):

(7)We are done. $A$ is now in REF.

## Example 2

**Use elementary row operations to put matrix $A = \begin{bmatrix} 2 & 1\\ 8 & 4 \end{bmatrix}$ into REF:**

First we will multiply row $1$ by $\frac{1}{2}$ ($\frac{1}{2} R_1 \to R_1$):

(8)Now let's take row $2$ and subtract $8$ times row $1$ ($R_2 - 8R_1 \to R_2$):

(9)We are done. $A$ is now in REF.