Definition

A matrix is in Row Echelon Form (REF) if

All zero rows are below all nonzero rows.
The leading entry of of each row to be to the right of the leading entry of the row above.

Example 1

For example, this matrix is in Row Echelon Form:

$$ \begin{align*} \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} $$

Definition

A matrix is in Reduced Row Echelon Form (RREF) if it is in REF and it satisfies the following two additional properties: c) The leading entries are all 1. d) The leading entries are the only non-zero entries in their column

The leading entries are all 1..
The leading entries are the only non-zero entries in their column.

Example 2

The same matrix from example 1 is not in RREF because column 2 for example still has non-zero entries above its leading entry.

$$ \begin{align*} \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} $$

Theorem 2

Every matrix can be put in RREF by a finite sequence of elementary row operations.

$R_i \leftrightarrow R_j $
$cR_i$ where $c \neq 0$
$R_i \rightarrow R_i + cR_j$

Proof: by construction (Gaussian Elimination) … TODO?

Theorem 3

The reduced RREF form of a matrix is unique. (Note row echelon form is not unique).

Theorem 4

The solution set of a linear system with augmented matrix in reduced row echelon form is easily described in a standard way.

Example 3

Consider the system with augmented matrix as follows:

$$ \begin{align*} \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \end{align*} $$

The last row has a single non-zero entry in the last column. This tells us that this system will not have a solution. To see this, translate back the system of linear equations

$$ \begin{array}{ll} x_1 \quad -x_3 &=0 \\ \quad x_2 &=0 \\ \quad \quad \quad \quad 0 &= 1 \end{array} $$

We can see here that $0=1$ has no solutions.

Example 4

Consider the system with augmented matrix as follows:

$$ \begin{align*} \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} $$

The last row has has all zeros. This tells us that we will have infinitely many solutions. To see this, translate back the system of linear equations

$$ \begin{array}{ll} x_1 \quad -x_3 &=0 \\ \quad x_2 &=0 \\ \quad \quad \quad \quad 0 &= 0 \end{array} $$

This time we have infinitely many solutions. The convention is to look for a column without a leading entry like Column 3. The convention is to use the third variable $x_3$ to parameterize the solution set. so set $x_3 = t$, then

$$ \begin{align*} x_3 &= t \\ x_1 &= t \\ x_2 &= 0. \end{align*} $$

The solution set is $(x_1,x_2,x_3) = \{(t, t, 0) | t \in \mathbf{R}\}$.

Example 5

Consider the system with augmented matrix as follows:

$$ \begin{align*} \begin{pmatrix} 1 & 2 & 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1 & 4 \\ \end{pmatrix}. \end{align*} $$

This is a system of three equations in 5 unknowns. Column one is $x_1$, column two is $x_2$ and so on. This matrix is also in Row Reduced Echelon Form because the leading entries are 1 and the leading entries are the only non-zero entries in their columns. Also, we don’t have a leading entry in the very last column so we do have a solution. This translates to the system of linear:

$$ \begin{array}{ll} x_1 + 2x_2 \quad \quad + x_4 &= 2 \\ \ \ \ \ \ \ \ \quad \quad \quad x_3 + 3x_4 &= 3 \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad x_5 &= 4 \end{array} $$

We have infinitely many solutions. So we’ll go by the convention which is to look for the columns without a leading entries and assign them variables. Here, columns 2 and 4 have no leading entries so set $x_2 = t_1$ and $x_4 = t_2$. This means that $x_3 = 3 - 3t_2$ and $x_1 = 2 - 2t_1 - t_2$. Writing this a solution vector:

$$ \begin{align*} \{(2-2t_1-t_2, t_1, 3-3t_2, t_2, 4) | t_1, t_2 \in \mathbf{R} \}. \end{align*} $$

Corollary

a system of linear equations has no solutions, exactly one solution (no parameters), or infinitely many solutions.

References:

Video Lectures from Math416 by Ely Kerman.