RREF

49. RREF

A matrix is in Reduced Row Echelon Form (RREF) if it satisfies all of the following:

Echelon form — see REF:
- Any all-zero rows are at the bottom
- Leading entries (pivots) move strictly right as you go down
Each pivot is $1$ (the leading entry of every non-zero row)
Each pivot is the only non-zero entry in its column (zeros above and below)

Example

RREF:

[\begin{matrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & - 2 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 \end{matrix}]

Not RREF (the pivot in column 2 is $3$ , not $1$ , and column 1 has a non-zero above the pivot):

[\begin{matrix} 1 & 5 & 2 \\ 0 & 3 & 4 \\ 0 & 0 & 0 \end{matrix}]

49.1. RREF vs REF

| Form | Pivots | Above pivots | |—|—|—| | REF | non-zero | unrestricted | | RREF | exactly $1$ | all $0$ |

REF comes from Gaussian elimination RREF is what Gauss–Jordan elimination produces.

49.2. Uniqueness

Given a matrix $𝐴$ , its REF is not unique (depends on row operations chosen), but its RREF is unique. So RREF is a canonical form — useful for:

Reading off the rank: number of non-zero rows
Reading off the solution to $𝐴 𝑥 = 𝑏$ directly from $[𝐴 | 𝑏]$ in RREF
Identifying linearly independent columns: the pivot columns
Computing the null space basis from free variables

49.3. Reading the solution

After reducing augmented matrix $[𝐴 | \vec{𝑏}]$ to RREF:

Example

[\begin{matrix} 1 & 0 & 3 & 0 & 5 & 4 \\ 0 & 1 & - 2 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1 & 6 & 2 \end{matrix}]

Pivot columns: 1, 2, 4 → pivot variables $𝑥_{1}, 𝑥_{2}, 𝑥_{4}$ . Free columns: 3, 5 → free variables $𝑥_{3}, 𝑥_{5}$ .

Solution (parametric):

\begin{matrix} 𝑥_{1} = 4 - 3 𝑥_{3} - 5 𝑥_{5} \\ 𝑥_{2} = 7 + 2 𝑥_{3} - 𝑥_{5} \\ 𝑥_{4} = 2 - 6 𝑥_{5} \\ 𝑥_{3}, 𝑥_{5} free \end{matrix}