Reduced Row Echelon Form

Echelon form is good enough for back substitution, but it is not unique: different sequences of row operations give different staircases. One more round of simplifying gives a form that is unique and lets you read the solution with no back substitution at all.

A matrix is in reduced row echelon form (RREF) when it is in echelon form and also (3) every pivot is $1$, and (4) each pivot is the only nonzero entry in its column. For example, $$\begin{bmatrix} 1 & 0 & 3 & 0 & -2 \\ 0 & 1 & -1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{bmatrix}$$ is in RREF: columns 1, 2 and 4 hold the pivots, and each of those columns is a single $1$ with zeros above and below.

Gauss–Jordan elimination

First reach echelon form as before (the forward phase). Then, starting with the rightmost pivot and moving left, scale each pivot row so the pivot is $1$ and use replacements to clear the entries above it (the backward phase).

Why it works

Why go right to left? When you clear above the pivot in column $j$, you add multiples of its row to the rows above. That row has zeros in every column to the left of $j$, so it cannot disturb the pivots already sitting in those columns. And the payoff: in RREF, each nonzero row contains exactly one pivot unknown, so each equation reads "pivot unknown $=$ number (minus free terms)". The fact that every matrix has exactly one RREF, whatever steps you take, is a theorem we will use often; it means the RREF is a fingerprint of the matrix.

Worked example: From a system to its RREF

Find the RREF of $\left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 2 & 5 & 4 & 11 \\ -1 & -1 & 2 & 1 \end{array}\right]$.

Forward: $R_{2} \to R_{2} - 2R_{1}$ and $R_{3} \to R_{3} + R_{1}$ clear column 1: $\left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 3 & 5 \end{array}\right]$
$R_{3} \to R_{3} - R_{2}$ clears below the second pivot: $\left[\begin{array}{ccc|c} 1 & 2 & 1 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \end{array}\right]$. Echelon form, with pivots $1, 1, 1$.
Backward, starting at the right: $R_{1} \to R_{1} - R_{3}$ and $R_{2} \to R_{2} - 2R_{3}$ clear above the third pivot: $\left[\begin{array}{ccc|c} 1 & 2 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$
$R_{1} \to R_{1} - 2R_{2}$ clears above the second pivot: $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$
Read the solution: $x = 4$, $y = -1$, $z = 2$. Check in row 2 of the original: $2(4) + 5(-1) + 4(2) = 11$. ✓

Answer: $\left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]$

Worked example: Scaling a pivot to 1

Find the RREF of $\begin{bmatrix} 2 & 4 & -2 & 6 \\ 1 & 3 & 2 & 7 \end{bmatrix}$.

$R_{1} \to \frac{1}{2}R_{1}$ makes the first pivot $1$: $\begin{bmatrix} 1 & 2 & -1 & 3 \\ 1 & 3 & 2 & 7 \end{bmatrix}$
$R_{2} \to R_{2} - R_{1}$: $\begin{bmatrix} 1 & 2 & -1 & 3 \\ 0 & 1 & 3 & 4 \end{bmatrix}$. The second pivot is already $1$.
$R_{1} \to R_{1} - 2R_{2}$ clears above it: $\begin{bmatrix} 1 & 0 & -7 & -5 \\ 0 & 1 & 3 & 4 \end{bmatrix}$

Answer: $\begin{bmatrix} 1 & 0 & -7 & -5 \\ 0 & 1 & 3 & 4 \end{bmatrix}$

Gauss–Jordan elimination, one step at a time. This system has a free variable, so no single point is shown.

Software uses RREF for small problems; for big ones it stops at echelon form, because the backward phase costs extra arithmetic. By hand, RREF is worth it when you need to describe all solutions.

Key idea

RREF is echelon form plus two rules: every pivot is $1$, and every pivot is alone in its column. Each matrix has exactly one RREF, and from it you can read the solutions directly.

Reduced Row Echelon Form

Gauss–Jordan elimination

Practice

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Sources and further reading