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MathidiaMathematics Academy
Numerical algorithms
Lesson 66 of 75
Introduction

Gaussian elimination

Objective. Reduce augmented matrices with controlled row operations.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Gaussian elimination solves a system of linear equations by writing them as an augmented matrix and applying row operations—swapping rows, multiplying a row by a non-zero constant, or adding a multiple of one row to another—to reduce the matrix to row-echelon form, where each row has more leading zeros than the one above. Back-substitution then solves for the unknowns starting from the simplest (last) equation and working upwards. For example, reducing a 3×3 system systematically eliminates one variable per row, leaving a triangular system that is solved step by step from the bottom.

Worked example

Use Gaussian elimination to solve the system x + y + z = 6, 2y + 5z = −4, 2x + 5y − z = 27.

6 … 2
Illustrated card showing 6 … 2
  1. 1Eliminate x from the third equation: R3 − 2R1 gives 2x+5y−z−2(x+y+z) = 27−12, so 3y−3z=15, i.e. y−z=5
  2. 2From the second equation, 2y+5z=−4; substitute y=z+5: 2(z+5)+5z=−4, so 7z+10=−4, giving 7z=−14, z=−2
  3. 3Then y = z+5 = 3
  4. 4Substitute into the first equation: x+3+(−2)=6, so x=5

Answer: x = 5, y = 3, z = −2

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Common mistake. Students perform row operations inconsistently, such as adding a multiple of one row to another without updating every entry in that row, which corrupts the solution. Always apply the chosen operation to every entry of the row, including the right-hand side constants.

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