Gaussian elimination
Objective. Solve and classify systems using row echelon form.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Gaussian elimination solves a system of linear equations by performing row operations (swap rows, scale a row, add a multiple of one row to another) on the augmented matrix to reach row echelon form, then uses back-substitution to find the solution. The resulting pivots reveal whether a system has a unique solution, infinitely many solutions, or none, depending on whether every variable has a pivot and whether any row reduces to a contradiction like 0=5. For the system x+y=3, 2x+y=4, subtracting twice the first row from the second gives −y=−2, so y=2, then x=1. Row echelon form must have leading entries strictly to the right as you move down the rows.
Worked example
Use Gaussian elimination to solve the system x+y=3, 2x+y=4.
- 1Write the augmented matrix [[1,1,|,3],[2,1,|,4]] and eliminate the first entry of row 2.
- 2Apply R2→R2−2R1: row 2 becomes [0,−1,|,−2].
- 3Solve the reduced equation −y=−2, giving y=2.
- 4Back-substitute into x+y=3: x=3−2=1.
Answer: x = 1, y = 2
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Start practisingCommon mistake. Students perform a row operation like R2→R2−2R1 but only update some entries of R2, forgetting to apply the operation to every entry in the row including the constant term on the right-hand side.
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