Laplace expansion

This article is about the expansion of the determinant of a square matrix as a weighted sum of determinants of sub-matrices. For the expansion of an 1/r-potential using spherical harmonical functions, see Laplace expansion (potential).

In linear algebra, the Laplace expansion, named after Pierre-Simon Laplace, also called cofactor expansion, is an expression for the determinant |B| of an n × n square matrix B that is a weighted sum of the determinants of n sub-matrices of B, each of size (n–1) × (n–1). 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.

The i, j cofactor of B is the scalar C_ij defined by

$C_{ij}\ = (-1)^{i+j} M_{ij}\,,$

where M_ij is the i, j minor matrix of B, that is, the determinant of the (n–1) × (n–1) matrix that results from deleting the i-th row and the j-th column of B.

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 its determinant |B| is given by:

$\begin{align}|B| & {} = b_{i1} C_{i1} + b_{i2} C_{i2} + \cdots + b_{in} C_{in} \\ & {} = b_{1j} C_{1j} + b_{2j} C_{2j} + \cdots + b_{nj} C_{nj}. \end{align}$

Examples

Consider the matrix

$B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}.$

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

$|B| = 1 \cdot \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

${} = 1 \cdot (-3) - 2 \cdot (-6) + 3 \cdot (-3) = 0.$

Alternatively, Laplace expansion along the second column yields

$|B| = -2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} - 8 \cdot \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix}$

${} = -2 \cdot (-6) + 5 \cdot (-12) - 8 \cdot (-6) = 0.$

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 $i,j\in\{1,2,\dots,n\}.$ For clarity we also label the entries of B that compose its i,j minor matrix M_ij as

(a_st) for $1 \le s,t \le n-1.$

Consider the terms in the expansion of | B | that have b_ij as a factor. Each has the form

$\sgn \tau\,b_{1,\tau(1)} \cdots b_{i,j} \cdots b_{n,\tau(n)} = \sgn \tau\,b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}$

for some permutation τ ∈ S_n with τ(i) = j, and a unique and evidently related permutation $\sigma\in S_{n-1}$ which selects the same minor entries as τ. Similarly each choice of σ determines a corresponding τ, i.e. the correspondence $\sigma\leftrightarrow\tau$ is a bijection between S_{n − 1} and $\{\tau\in S_n\colon\tau(i)=j\}.$ The permutation τ can be derived from σ as follows.

Define $\sigma'\in S_n$ by σ'(k) = σ(k) for $1 \le k \le n-1$ and σ'(n) = n. Then sgn σ' = sgn σ and

$\tau\,=(j,j+1,\ldots,n)\sigma'(n,n-1,\ldots,i)$

Since the two cycles can be written respectively as n − j and n − i transpositions,

$\sgn\tau\,= (-1)^{2n-(i+j)} \sgn\sigma'\,= (-1)^{i+j} \sgn\sigma.$

And since the map $\sigma\leftrightarrow\tau$ is bijective,

$\sum_{\tau \in S_n\colon\tau(i)=j}$	$\sgn \tau\,b_{1,\tau(1)} \cdots b_{n,\tau(n)}$
	$= \sum_{\sigma \in S_{n-1}} (-1)^{i+j}\sgn\sigma\, b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}$
	$=\ b_{ij} (-1)^{i+j} |M_{ij}|,$

from which the result follows.

References

• David Poole: Linear Algebra. A Modern Introduction. Cengage Learning 2005, ISBN 0534998453, p. 265-267 (restricted online copy at Google Books)
• Harvey E. Rose: Linear Algebra. A Pure Mathematical Approach. Springer 2002, ISBN 3764369051, p. 57-60 (restricted online copy at Google Books)

See also

• Leibniz formula for determinants

External links

• Laplace expansion at PlanetMath