行列式に対するライプニッツの明示公式
, n i ≠ j 1 , i ≠ j 2 a σ ( i ) i ) {\displaystyle F(\mathbf {a} ^{1},\dotsc ,\mathbf {a} ^{j_{1}},\dotsc ,\mathbf {a} ^{j_{2}},\dotsc ,\mathbf {a} ^{n})=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\alpha _{\sigma }a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}\qquad {\Bigl (}\alpha _{\sigma }:=\prod _{i=1,\dotsc ,n \atop i\neq j_{1},i\neq j_{2}}a_{\sigma (i)}^{i}{\Bigr )}} において、各 σ ∈ Sn に対し、σ から添字 j1 と j2 を入れ替えて得られる置換を σ′ と書くことにすれば、右辺はさらに ∑ σ ∈ S n , σ ( j 1 ) < σ ( j 2 ) ( sgn  ( σ ) α σ a σ ( j 1 ) j 1 a σ ( j 2 ) j 2 + sgn  ( σ ′ ) α σ ′ a σ ′ ( j 1 ) j 1 a σ ′ ( j 2 ) j 2 ) = ∑ σ ∈ S n σ ( j 1 ) < σ ( j 2 ) ( sgn  ( σ ) α σ a σ ( j 1 ) j 1 a σ ( j 2 ) j 2 − sgn  ( σ ) α σ a σ ( j 2 ) j 1 a σ ( j 1 ) j 2 ) = ∑ σ ∈ S n σ ( j 1 ) < σ ( j 2 ) sgn  ( σ ) α σ ( a σ ( j 1 ) j 1 a σ ( j 2 ) j 2 − a σ ( j 1 ) j 2 a σ ( j 2 ) j 1 ) {\displaystyle {\begin{aligned}&\quad \sum _{\sigma \in S_{n}, \atop \sigma (j_{1})<\sigma (j_{2})}{\Bigl (}\operatorname {sgn}(\sigma )\alpha _{\sigma }a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}+\operatorname {sgn}(\sigma ')\alpha _{\sigma '}a_{\sigma '(j_{1})}^{j_{1}}a_{\sigma '(j_{2})}^{j_{2}}{\Bigr )}\\&=\sum _{\sigma \in S_{n} \atop \sigma (j_{1})<\sigma (j_{2})}{\bigl (}\operatorname {sgn}(\sigma )\alpha _{\sigma }a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}-\operatorname {sgn}(\sigma )\alpha _{\sigma }a_{\sigma (j_{2})}^{j_{1}}a_{\sigma (j_{1})}^{j_{2}}{\bigr )}\\&=\sum _{\sigma \in S_{n} \atop \sigma (j_{1})<\sigma (j_{2})}\operatorname {sgn}(\sigma )\alpha _{\sigma }{\bigl (}a_{\sigma (j_{1})}^{j_{1}}a_{\sigma (j_{2})}^{j_{2}}-a_{\sigma (j_{1})}^{j_{2}}a_{\sigma (j_{2})}^{j_{_{1}}}{\bigr )}\end{aligned}}} と書き直せるから、 a j 1 = a j 2 ⟹ F ( A ) = 0 {\displaystyle \mathbf {a} ^{j_{1}}=\mathbf {a} ^{j_{2}}\implies F(A)=0} を得る。