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)

\begin{align} \quad A = \begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 2 \end{bmatrix} \quad , \quad B = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \quad , \quad C = \begin{bmatrix} 1 &0 \\ 0 & 1\\ 0 & 0\\ 0 & 0\\ 0 & 0 \end{bmatrix} \quad , \quad D = \begin{bmatrix} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0\\ 0 & 0 & 1 & 3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \end{align}

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)

\begin{bmatrix} 1 & \frac{1}{3}\\ 3 & 4 \end{bmatrix}

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

(3)

\begin{bmatrix} 1 & \frac{1}{3}\\ 1 & \frac{4}{3} \end{bmatrix}

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

(4)

\begin{bmatrix} 1 & \frac{1}{3}\\ 0 & 1 \end{bmatrix}

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)

\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}

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)

\begin{bmatrix} 1 & 2& 3 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 &1 & 1 \end{bmatrix}

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

(7)

\begin{bmatrix} 1 & 2& 0 & 0 & 3 & 0\\ 0 & 0 & 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 &1 & 1 \end{bmatrix}

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