In linear algebra, the Laplace expansion, named after Pierre-Simon Laplace, also called cofactor expansion, is an expression of the determinant of an n × n -matrix B as a weighted sum of minors, which are the determinants of some (n − 1) × (n − 1) -submatrices of B. Specifically, for every i, the Laplace expansion along the ith row is the equality where b_{i,j} is the entry of the ith row and jth column of B, and m_{i,j} is the determinant of the submatrix obtained by removing the ith row and the jth column of B. Similarly, the Laplace expansion along the jth column is the equality (Each identity implies the other, since the determinants of a matrix and its transpose are the same.) The coefficient of b_{i,j} in the above sum is called the cofactor of b_{i,j} in B. The Laplace expansion is often useful in proofs, as in, for example, allowing recursion on the size of matrices. It is also of didactic interest for its simplicity and as one of several ways to view and compute the determinant. For large matrices, it quickly becomes inefficient to compute when compared to Gaussian elimination.

Examples

Consider the matrix The determinant of this matrix can be computed by using the Laplace expansion along any one of its rows or columns. For instance, an expansion along the first row yields: Laplace expansion along the second column yields the same result: It is easy to verify that the result is correct: the matrix is singular because the sum of its first and third column is twice the second column, and hence its determinant is zero.

Proof

Suppose B is an n × n matrix and For clarity we also label the entries of B that compose its i,j minor matrix M_{ij} as (a_{st}) for Consider the terms in the expansion of |B| that have b_{ij} as a factor. Each has the form for some permutation τ ∈ Sn with \tau(i)=j, and a unique and evidently related permutation which selects the same minor entries as τ. Similarly each choice of σ determines a corresponding τ i.e. the correspondence is a bijection between S_{n-1} and Using Cauchy's two-line notation, the explicit relation between \tau and \sigma can be written as where is a temporary shorthand notation for a cycle. This operation decrements all indices larger than j so that every index fits in the set {1,2,...,n-1} The permutation τ can be derived from σ as follows. Define by for and. Then \sigma' is expressed as Now, the operation which apply first and then apply \sigma' is (Notice applying A before B is equivalent to applying inverse of A to the upper row of B in two-line notation) where is temporary shorthand notation for. the operation which applies \tau first and then applies is above two are equal thus, where is the inverse of which is. Thus Since the two cycles can be written respectively as n-i and n-j transpositions, And since the map is bijective, from which the result follows. Similarly, the result holds if the index of the outer summation was replaced with j.

Laplace expansion of a determinant by complementary minors

Laplace's cofactor expansion can be generalised as follows.

Example

Consider the matrix The determinant of this matrix can be computed by using the Laplace's cofactor expansion along the first two rows as follows. Firstly note that there are 6 sets of two distinct numbers in {1, 2, 3, 4}, namely let be the aforementioned set. By defining the complementary cofactors to be and the sign of their permutation to be The determinant of A can be written out as where H^{\prime} is the complementary set to H. In our explicit example this gives us As above, it is easy to verify that the result is correct: the matrix is singular because the sum of its first and third column is twice the second column, and hence its determinant is zero.

General statement

Let B=[b_{ij}] be an n × n matrix and S the set of k -element subsets of , H an element in it. Then the determinant of B can be expanded along the k rows identified by H as follows: where is the sign of the permutation determined by H and L, equal to, b_{H,L} the square minor of B obtained by deleting from B rows and columns with indices in H and L respectively, and c_{H,L} (called the complement of b_{H,L}) defined to be b_{H',L'} , H' and L' being the complement of H and L respectively. This coincides with the theorem above when k=1. The same thing holds for any fixed k columns.

Computational expense

The Laplace expansion is computationally inefficient for high-dimension matrices, with a time complexity in big O notation of O(n!) . Alternatively, using a decomposition into triangular matrices as in the LU decomposition can yield determinants with a time complexity of O(n3) . The following Python code implements the Laplace expansion: