Determinante Matriz 3×3

3×3 determinant: Sarrus rule and cofactor expansion

Take a 3×3 matrix [[a, b, c], [d, e, f], [g, h, i]]. The Sarrus rule gives you det = aei + bfg + cdh - ceg - afh - bdi, which is just the three left-to-right diagonals minus the three right-to-left ones. You get the same answer from a Laplace expansion along the first row, det = a(ei - fh) - b(di - fg) + c(dh - eg), where each entry is multiplied by its signed 2×2 cofactor. Worked out: det([[1, 2, 3], [4, 5, 6], [7, 8, 10]]) = 1·(50 - 48) - 2·(40 - 42) + 3·(32 - 35) = 2 + 4 - 9 = -3. With det ≠ 0, the matrix is invertible. Picture it geometrically and |det| is the volume of the parallelepiped spanned by the three row (or column) vectors; the sign tells you the orientation, with positive keeping the handedness and negative flipping it.

Applications

It shows up when you solve 3×3 linear systems with Cramer's rule, when you check whether a 3D linear transformation (rotation, scaling, shear) is invertible in computer graphics and robotics, and when you need the volume of a parallelepiped V = |det|. It's also the Jacobian in triple integrals when you switch to spherical or cylindrical coordinates, and it gives you cross products through (u × v)·w = det([u, v, w]), the scalar triple product.

FAQ

Does the Sarrus rule work for 4×4 matrices? No, it only works for 3×3. Once you hit 4×4 or bigger you'll reach for Laplace expansion, or in practice Gaussian elimination (O(n³)).

Why does det = 0 mean the three vectors are coplanar? The parallelepiped flattens out to zero volume, which can only happen when all three vectors sit in the same plane and are linearly dependent.

Is the determinant invariant under row swaps? No. Swap two rows and the sign flips. If you multiply a row by k, det gets multiplied by k too. Adding a multiple of one row to another, though, leaves det exactly as it was.