Beyond Row Echelon Form

Linear Algebra (Matrices) · Axiom Academy

LESSON Beyond Row Echelon Form Reduced row echelon form: scale every pivot to 1, clear above it too, and read the solution straight off the page. Start from the augmented matrix already in row echelon form . The pivots march down a staircase and everything below them is zero — but the pivots are still 2 , 5 , and 3 (not 1 ), and there are non-zero entries sitting above them. REF stops here; to solve, you would still have to back-substitute. Pivots (leading entries) trace the staircase Any all-zero rows sit at the bottom Each pivot is strictly to the right of the pivot above it Every entry below a pivot is zero 2. Step One of RREF: Make Every Pivot a 1 The first new requirement: each pivot must be a leading 1 . Divide each pivot row by its own pivot value — row 1 by 2 , row 2 by 5 , row 3 by 3 . Dividing a whole equation by a constant never changes its solution, so the system is untouched; only the numbers get cleaner. Multiply the entire row (including its augmented entry) by the reciprocal of its pivot. Scaling an equation by a non-zero constant is a legal row operation — the solution set is identical. 3. Step Two of RREF: Clear Above Each Pivot The second new requirement: each pivot column holds a single 1 with zeros above and below . Working from the bottom pivot upward, subtract the right multiple of each pivot row from the rows above it, zeroing every entry over a pivot. When the sweep finishes, the coefficient block is a clean identity matrix — this is RREF .

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