The Gauss-Jordan Algorithm

Linear Algebra (Matrices) · Axiom Academy

EXAMPLE The Gauss-Jordan Algorithm A full Gauss-Jordan reduction of one augmented matrix — forward eliminate, scale each pivot to 1, clear above and below, read the answer off the identity block. Solve this system of linear equations using Gauss-Jordan elimination — reduce its augmented matrix to reduced row echelon form (RREF) , then read the solution directly. The target: a 1 on each diagonal, 0 everywhere else — then the right column is the solution. Nicely done — you ran a complete Gauss-Jordan reduction all the way to the identity block. Here is what you practiced: Forward pass: Work left to right. For each pivot column, make the pivot a 1 by scaling its row, then use that row to create 0 s below it. Fractions are fine: Scaling a pivot to 1 often produces fractions like — that is expected; carry them exactly rather than rounding. Backward pass: Once you reach row echelon form, work from the bottom-right pivot upward, clearing the 0 s above each pivot to reach RREF. The whole point of RREF: With a 1 on every diagonal and 0 s elsewhere, each row just says x= , y= , z= — the right-hand column is the solution. Result: . Substitute back to check: 1 + 2(2) + 3 = 8 ✓ ︎ , 2(1) + 2(2) - 3 = 3 ✓ ︎ , -1 + 2 + 2(3) = 7 ✓ ︎ . Gauss-Jordan is mechanical and never gets stuck: pick the pivot, scale it to 1 , clear the column. Repeat for every pivot and any consistent system with a unique solution falls right out of the matrix.

This is the written version of the interactive lesson above. See the full Linear Algebra (Matrices) course.