Gaussian elimination and row echelon form
Reduce a system with row swaps, nonzero row scaling and row replacement to echelon or reduced echelon form, then read pivots, free variables and the full solution set.
What a learner can do afterwards
- Carry a system to row echelon form and recognize reduced row echelon form
- Identify pivot and free columns and write the general solution
- Say why each row operation preserves the solution set
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Elimination uses three moves: swap two rows, scale a row by a nonzero number, or add a multiple of one row to another. Each move is reversible, so the solution set never changes. Each matrix row is an equation. Columns before the bar hold coefficients; the bar separates the constants. Work top to bottom toward echelon form, also called staircase form, then back substitute from the last pivot up.
Solve x + y = 5 and x - y = 1. Add the rows and y cancels: 2x = 6, so x = 3. Then y = 2, and both equations hold. Adding was the allowed move of adding one row to another.
In echelon form, zero rows come last. Each pivot is the first nonzero entry of a nonzero row and lies to the right of the pivot above. Reduced form also makes each pivot 1 and the only nonzero entry in its column. In [1 0 2 | 3; 0 1 -1 | 4; 0 0 0 | 0], coefficient columns 1 and 2 have pivots; column 3 is free. The last column holds constants, not a variable. For a consistent system, free variables equal unknowns minus pivots: 3 minus 2 gives 1 here.
The augmented matrix of x + y + z = 5, 2x + 2y + 2z = 10 and x - y = 1, with rows R1, R2, R3, is [1 1 1 | 5; 2 2 2 | 10; 1 -1 0 | 1]. R2 <- R2 - 2R1 gives [0 0 0 | 0]. R3 <- R3 - R1 gives [0 -2 -1 | -4]. Swap R2 and R3. Column 3 has no pivot, so z = t. Then -2y - t = -4 gives y = 2 - t/2, and R1 gives x = 3 - t/2. To clear an entry below a pivot, divide that entry by the pivot and subtract that multiple of the pivot row.
One move per cell. The general solution is (x, y, z) = (3 - t/2, 2 - t/2, t).
A row 0 = 5 is a contradiction, so no solution exists. A row 0 = 0 is redundant, but it does not by itself prove infinitely many solutions; the system must be consistent and have a free variable. Scaling by a nonzero number c is undone by scaling by 1/c, so scaling by 4 is undone by 1/4. Scaling a row by zero is not allowed because it loses information and cannot be reversed.
Swap, scale, and add toward staircase form, then read pivots, free columns, and the solution.
Original lesson and assessment text by the Open Source Taxonomy project.
The rewritten readings were written for this dataset. Includes a Khan Academy video and original practice questions. Individual source titles are listed with the resources. The source licenses for the remaining imported resources are unknown.
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