Laplace expansion

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 Cij defined by

C_{ij}\ = (-1)^{i%2Bj} M_{ij}\,,

where Mij 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 = (bij) 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} %2B b_{i2} C_{i2} %2B \cdots %2B b_{in} C_{in} \\ & {} = b_{1j} C_{1j} %2B b_{2j} C_{2j} %2B \cdots %2B b_{nj} C_{nj}. \end{align}

Contents

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} %2B 3 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}
{} = 1 \cdot (-3) - 2 \cdot (-6) %2B 3 \cdot (-3) = 0.

Alternatively, Laplace expansion along the second column yields

|B| = -2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} %2B 5 \cdot \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} - 8 \cdot \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix}
{} = -2 \cdot (-6) %2B 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 ∈ Sn with \tau(i)=j, and a unique and evidently related permutation \sigma\in S_{n-1} which selects the same minor entries as \tau. Similarly each choice of \sigma determines a corresponding \tau, 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 \tau can be derived from \sigma as follows.

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

\tau\,=(j,j%2B1,\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%2Bj)} \sgn\sigma'\,= (-1)^{i%2Bj} \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%2Bj}\sgn\sigma\, b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}
=\ b_{ij} (-1)^{i%2Bj} |M_{ij}|,

from which the result follows.

References

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