Laplace expansion

In linear algebra, the Laplace expansion of the determinant of an n × n square matrix B expresses the determinant |B| as a sum of n determinants of (n-1) × (n-1) sub-matrices of B. There are n² such expressions, one for each row and column of B. The Laplace expansion is of theoretical interest as one of several ways to view the determinant, as well as of practical use in determinant computation.

Define the i,j minor matrix M_ij of B as the (n-1) × (n-1) matrix that results from deleting the i-th row and the j-th column of B, and C_i,j the cofactor of B as

Then the Laplace expansion is given by the following

Theorem Suppose B = (b_ij) is an n × n matrix and i,j ∈ {1, 2, ...,n}.

Then the determinant

Examples

Consider the matrix

The determinant of this matrix can be computed by using the Laplace expansion along the first row:

Alternatively, Laplace expansion along the second column yields

It is easy to see 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 ∈ S_n with τ(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 The permutation τ can be derived from σ as follows.

Define by σ'(k) = σ(k) for and σ'(n) = n. Then sgnσ' = sgnσ and

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.

See also

• Laplace expansion (potential)