How to find RREF by hand: Gauss-Jordan elimination

Gauss-Jordan elimination is the process for finding reduced row echelon form. It's systematic: you work column by column, left to right, creating a pivot in each column and then zeroing out everything else in that column, above and below.

The three tools you have are the three elementary row operations:

  1. Swap two rows: R1 ↔ R2
  2. Multiply a row by a nonzero scalar: R2 → (1/3)·R2
  3. Add a multiple of one row to another: R3 → R3 − 4·R1

Every step in Gauss-Jordan is one of these three operations. Nothing else is allowed.

The example

We'll reduce this augmented matrix for a system of 3 equations in 3 unknowns:

[ 2   4  -2  |  2 ]
[ 4   9  -3  |  8 ]
[-2  -3   7  | 10 ]

The vertical bar separates the coefficient columns from the constants (the right-hand side of the equations). Row operations apply to the entire row, bar included.

Step 1: Get a pivot of 1 in position (1,1)

Multiply row 1 by 1/2:

R1 → (1/2)·R1

[ 1   2  -1  |  1 ]
[ 4   9  -3  |  8 ]
[-2  -3   7  | 10 ]

Step 2: Zero out column 1 below the pivot

Eliminate the 4 in row 2 and the −2 in row 3.

R2 → R2 − 4·R1

R3 → R3 + 2·R1

[ 1   2  -1  |  1 ]
[ 0   1   1  |  4 ]
[ 0   1   5  | 12 ]

Row 2: [4 − 4·1, 9 − 4·2, −3 − 4·(−1), 8 − 4·1] = [0, 1, 1, 4]. Correct.

Row 3: [−2 + 2·1, −3 + 2·2, 7 + 2·(−1), 10 + 2·1] = [0, 1, 5, 12]. Correct.

Step 3: The pivot in position (2,2) is already 1

No scaling needed. Now zero out column 2, both above (row 1) and below (row 3).

R1 → R1 − 2·R2

R3 → R3 − R2

[ 1   0  -3  | -7 ]
[ 0   1   1  |  4 ]
[ 0   0   4  |  8 ]

Row 1: [1 − 0, 2 − 2·1, −1 − 2·1, 1 − 2·4] = [1, 0, −3, −7]. Correct.

Row 3: [0 − 0, 1 − 1, 5 − 1, 12 − 4] = [0, 0, 4, 8]. Correct.

Step 4: Get a pivot of 1 in position (3,3)

R3 → (1/4)·R3

[ 1   0  -3  | -7 ]
[ 0   1   1  |  4 ]
[ 0   0   1  |  2 ]

Step 5: Zero out column 3 above the pivot

R1 → R1 + 3·R3

R2 → R2 − R3

[ 1   0   0  | -1 ]
[ 0   1   0  |  2 ]
[ 0   0   1  |  2 ]

Row 1: [1 + 0, 0 + 0, −3 + 3·1, −7 + 3·2] = [1, 0, 0, −1]. Correct.

Row 2: [0 − 0, 1 − 0, 1 − 1, 4 − 2] = [0, 1, 0, 2]. Correct.

Reading the solution

The matrix is now in RREF. The left 3×3 block is the identity matrix, and the solution reads directly from the right column:

x₁ = −1, x₂ = 2, x₃ = 2

You can verify by substituting back into the original equations.

What to watch for

Fractions early. Multiplying a row by a reciprocal in step 1 sometimes produces awkward fractions through the rest of the work. An alternative: swap rows first to get a row with a leading 1 before you scale anything. In this example, there was no clean way to avoid the 1/2 scaling, but in other problems you might find a row that already has a leading 1.

Arithmetic errors compound. A mistake in step 2 propagates through every subsequent step. If your final matrix doesn't look like RREF, go back and check each row operation individually rather than searching the final answer.

Column order matters. Always work left to right. If a column has no nonzero entry to use as a pivot (all zeros below the current position), skip it and move to the next column. That column corresponds to a free variable, which comes up when a system has infinitely many solutions.

If you want to check your work or skip straight to the answer, the RREF calculator computes the full reduction using exact fraction arithmetic, so there's no floating-point rounding.