Laplace Expansion - Proof

Proof

Suppose is an n × n matrix and For clarity we also label the entries of that compose its minor matrix as

for

Consider the terms in the expansion of that have 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, and a unique and evidently related permutation which selects the same minor entries as Similarly each choice of determines a corresponding i.e. the correspondence is a bijection between and The permutation can be derived from as follows.

Define by for and . Then and

Since the two cycles can be written respectively as and transpositions,

And since the map is bijective,

= \sum_{\sigma \in S_{n-1}} (-1)^{i+j}\sgn\sigma\, b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}

from which the result follows.

Read more about this topic:  Laplace Expansion

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