Calculating determinants with Gaussian elimination
For 3×3 and 4×4 matrices Gaussian elimination is often the fastest way: bring the matrix to triangular form, multiply the diagonal and keep track of what the row operations did to the determinant.
How to do it
- 1
Bring the matrix to upper triangular form (zeros below the diagonal) with row operations.
- 2
Adding a multiple of one row to another does not change the determinant.
- 3
Every row swap flips the sign.
- 4
Multiplying a row by k multiplies the determinant by k as well; divide it out at the end.
- 5
The determinant of a triangular matrix is the product of its diagonal entries.
Example with worked solution
Worked solution
Starting matrix
6 1 0 −3 −1 1 3 −1 6 1. II + 1/2·I
6 1 0 0 −1/2 1 3 −1 6 2. III − 1/2·I
6 1 0 0 −1/2 1 0 −3/2 6 3. III − 3·II
6 1 0 0 −1/2 1 0 0 3
Echelon form: det = 6 · (−1/2) · 3 = −9
Your turn
Enter the row operations, e.g. II − 3·I. If you get stuck, "Hint: next step" helps.
Calculate the determinant.
Swapping rows flips the sign; multiplying a row by k multiplies the determinant by k. Adding a multiple of another row changes nothing.
| 3 | 2 | 0 |
| −3 | −2 | −1 |
| −6 | −2 | −3 |
Your answer
Typical mistakes
- Not counting row swaps: then the sign is wrong.
- Using combinations like 2·II − 3·I without accounting for the factor 2: the determinant doubled.
- Applying the rule of Sarrus to a 4×4 matrix: it only works for 3×3 matrices.
Frequently asked questions
When is the determinant 0?
When the rows (or columns) are linearly dependent, for example when a zero row appears during elimination. The matrix is then not invertible and the linear system has no unique solution.
Sarrus, Laplace or Gauss?
Sarrus is quick for 3×3. If a row or column has many zeros, Laplace expansion pays off. From 4×4 on, Gaussian elimination is usually the least work.
What does the determinant mean geometrically?
Its absolute value is the factor by which the matrix scales areas (2×2) or volumes (3×3). The sign shows whether the orientation is kept.