Matrix 3x3 Determinant

3×3 determinant: Sarrus rule and Laplace expansion

For a 3×3 matrix [[a, b, c], [d, e, f], [g, h, i]] the Sarrus rule gives det = aei + bfg + cdh - ceg - afh - bdi: add up the three left-to-right diagonals and subtract the three right-to-left ones. You can also use Laplace cofactor expansion. Expanding along the first row gives det = a(ei - fh) - b(di - fg) + c(dh - eg), the same number written out as 2×2 sub-determinants. To see it in numbers: |[1, 2, 3]; [4, 5, 6]; [7, 8, 10]| = 1·(50 - 48) - 2·(40 - 42) + 3·(32 - 35) = 2 + 4 - 9 = -3. The same reading holds here as in any dimension. When det ≠ 0 the matrix is invertible and the three rows (or columns) are linearly independent; when det = 0 they are coplanar and the matrix is singular. Keep in mind that Sarrus only works for 3×3. For 4×4 or larger, fall back to Laplace or Gaussian elimination.

Applications

It drives Cramer's rule for 3×3 linear systems and gives the volume of the parallelepiped spanned by three 3D vectors (|det|). It's the Jacobian for changes of variable in triple integrals (spherical, cylindrical), an alternative way to get cross product magnitude, and it turns up across 3D computer graphics (orientation tests, ray-triangle intersection). Analytical geometry uses it too, for checking collinearity and coplanarity.

FAQ

Does Sarrus work for 4×4? No. The diagonal trick is a coincidence that happens to hold for 3×3 and nothing larger. From 4×4 onward, use Laplace or row reduction.

What does the sign of the determinant tell me? A positive value says the three vectors form a right-handed, positively oriented basis. Negative means left-handed. Either way, the magnitude is the volume, no matter the sign.

Is there a faster way for large matrices? Yes. Gaussian elimination computes det in O(n³): reduce to triangular form, then multiply the pivots. Laplace, by contrast, is O(n!).