# Reduced Row Echelon Form of a Matrix (RREF)

We’ve looked at what it means for a matrix to be in Row Echelon Form (REF). There is another form that a matrix can be in, known as **Reduced Row Echelon Form** (often abbreviated as RREF). This form is simply an extension to the REF form, and is very useful in solving systems of linear equations as the solutions to a linear system become a lot more obvious. We will now define what it means for an $m \times n$ matrix $A$ to be in this form while noticing that the only difference between RREF and REF is the addition of condition 4.

Definition: Let $A$ be an $m \times n$ matrix. Then $A$ is said to be in Reduced Row Echelon Form if $A$ satisfies the following four 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.4. Every column that contains a leading $1$ must have zeros everywhere else in that column. |

For example, the following matrices $A$, $B$, $C$ and $D$ are all in RREF.

(1)

Notice that how in all of the matrices above, if a column contains a leading 1 then the rest of the entries in that column are zeroes. That is essentially the only difference between REF and RREF. Like with turning a matrix into REF with elementary row operations, we can also do the same for RREF. The strategy is to first convert a matrix to REF and is best explained with an example.

## Example 1

**Use elementary row operations to take the following matrix $A = \begin{bmatrix} 3 & 1\\ 3 & 4 \end{bmatrix}$ and convert it into RREF:**

Our first step is to take row $1$ and multiply it by $\frac{1}{3}$ ($\frac{1}{3}R_1 \to R_1$):

(2)

Now let’s take row $2$ and multiply it by $\frac{1}{3}$ ($\frac{1}{3}R_2 \to R_2$):

(3)

Now take row $2$ and subtract row $1$ from it ($R_2 – R_1 \to R_2$):

(4)

This matrix is now in REF. To turn it into RREF, we will in a sense work backwards. We will take row $1$ and subtract $3$ times row $2$ ($R_1 – 3R_2 \to R_1$):

(5)

Hence we are done as the resulting matrix satisfies all four conditions to be in RREF.

## Example 2

**Take the follow matrix and put it in RREF: $\begin{bmatrix} 1 & 2& 3 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 &0 & 1\\ 0 & 0 & 0 & 1 &1 & 1 \end{bmatrix}$.**

We first note that this matrix is already in REF, so all we have to do it go from REF to RREF.

First let’s take row $2$ and subtract row $3$ ($R_2 – R_3 \to R_2$):

(6)

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

(7)

Hence we are done as the resulting matrix satisfies all four conditions to be in RREF.