Row Echelon Form (REF) and Reduced Row Echelon Form (RREF)

REF vs. RREF — definitions

A matrix is in row-echelon form (REF) if:

1. All zero rows sit at the bottom.
2. The leading entry (pivot) of each non-zero row is strictly to the right of the pivot in the row above.
3. Entries below each pivot are zero.

A matrix is in reduced row-echelon form (RREF) if, additionally: 4. Each pivot is exactly $1$. 5. Each pivot is the only non-zero entry in its column.

RREF is unique — any matrix reduces to exactly one RREF regardless of the row operations you use.

Step-by-step reduction

Start: $\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}$

Step 1 — Zero out column 1 below the pivot.

$R_2 \to R_2 - 4R_1$: $(4, 5, 6) - 4(1, 2, 3) = (0, -3, -6)$. $R_3 \to R_3 - 7R_1$: $(7, 8, 10) - 7(1, 2, 3) = (0, -6, -11)$.

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -11 \end{pmatrix}$

Step 2 — Zero out column 2 below the pivot.

$R_3 \to R_3 - 2 R_2$: $(0, -6, -11) - 2(0, -3, -6) = (0, 0, 1)$.

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 1 \end{pmatrix} \quad \text{(REF)}$

This is row-echelon form. To continue to RREF:

Step 3 — Scale each row so pivots become 1.

$R_2 \to -\tfrac{1}{3} R_2$: $(0, -3, -6) \to (0, 1, 2)$.

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}$

Step 4 — Zero out entries ABOVE each pivot. Start from the rightmost pivot.

$R_1 \to R_1 - 3 R_3$: $(1, 2, 3) - 3(0, 0, 1) = (1, 2, 0)$. $R_2 \to R_2 - 2 R_3$: $(0, 1, 2) - 2(0, 0, 1) = (0, 1, 0)$.

$\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

$R_1 \to R_1 - 2 R_2$: $(1, 2, 0) - 2(0, 1, 0) = (1, 0, 0)$.

$\boxed{\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}} \quad \text{(RREF)}$

The matrix has three pivots — one in each column — so its rank is 3.

What RREF tells you

• Rank = number of pivots.
• Number of free variables = (# columns) − (# pivots).
• Null space basis reads directly off the RREF of $[A \mid 0]$.
• Solving $A\mathbf{x} = \mathbf{b}$ reduces to reading values from the RREF of $[A \mid \mathbf{b}]$.

For this $3 \times 3$ identity RREF: the matrix is invertible, with a unique solution to any consistent system.

Common mistakes

• Moving column-to-column before clearing the current column. Clear one column completely before starting the next.
• Scaling before clearing the column. Works either way, but leave scaling for last to keep arithmetic clean (fewer fractions mid-way).
• Confusing REF with RREF. REF only requires zeros below pivots and pivots moving right. RREF also requires leading $1$s and zeros above each pivot.

Try it in the visualization

Step through each row operation. The matrix updates in place; pivots highlight in green; zero entries below and above pivots highlight as they're cleared.