Laplace Expansion

(A) = _j (-1)^i+j a_ij M_ij — expand along any row or column.

The idea

Theorem (Laplace (cofactor) expansion).

Let $A$ be an $n \times n$ matrix and fix any one row $i$. Then $\det A = \sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij},$ where the minor $M_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix left after deleting row $i$ and column $j$ from $A$. The identical formula holds down any column $j$, summing over $i$.

The formula reduces a determinant to smaller determinants. A $2 \times 2$ determinant has the direct rule $ad - bc$, so the expansion trades each $3 \times 3$ determinant for three $2 \times 2$ ones, each $4 \times 4$ for four $3 \times 3$ ones, and so on, until every piece can be computed directly.

Two features of the formula guide its use. The factor $(-1)^{i+j}$ is $+1$ when $i + j$ is even and $-1$ when it is odd, so the signs alternate in a checkerboard pattern that starts with $+$ in the top-left corner. And an entry $a_{ij}$ equal to $0$ contributes $0 \cdot M_{ij} = 0$ whatever its minor is, so that minor never has to be computed. Since every row and every column gives the same answer, expand along the one with the most zeros.

Algorithm.

Algorithm: Laplace Expansion Input: an n × n matrix A, n ≥ 2 Output: det A 1. if n = 2, return a11 a22 - a12 a21 // the direct 2 × 2 rule 2. pick the row or column with the most zeros 3. for each nonzero entry a_ij in it: delete row i and column j, get the minor M_ij by recursion 4. return Σ (-1)^(i+j) a_ij M_ij, summed over those entries // checkerboard signs; zero entries contribute nothing

Ways to work on it

• Walkthrough. Cofactor expansion of a 3 × 3 matrix along the first row.
• Practice. Compute a 3 × 3 determinant — the matrix has zeros, so pick wisely.
• Hardest. Dense 3 × 3 with negative entries — one shot.

Not sure where to start? Take the ten-question placement test.