エディントンのイプシロン
3×3行列式は
det [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] = det [ ∑ i δ 1 i a i 1 ∑ j δ 1 j a j 2 ∑ k δ 1 k a k 3 ∑ i δ 2 i a i 1 ∑ j δ 2 j a j 2 ∑ k δ 2 k a k 3 ∑ i δ 3 i a i 1 ∑ j δ 3 j a j 2 ∑ k δ 3 k a k 3 ] = ∑ i , j , k det [ δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 3 i δ 3 j δ 3 k ] a i 1 a j 2 a k 3 = ∑ i , j , k ε i j k a i 1 a j 2 a k 3 {\displaystyle {\begin{aligned}\det {\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}}&=\det {\begin{bmatrix}\sum _{i}\delta _{1i}a_{i1}&\sum _{j}\delta _{1j}a_{j2}&\sum _{k}\delta _{1k}a_{k3}\\\sum _{i}\delta _{2i}a_{i1}&\sum _{j}\delta _{2j}a_{j2}&\sum _{k}\delta _{2k}a_{k3}\\\sum _{i}\delta _{3i}a_{i1}&\sum _{j}\delta _{3j}a_{j2}&\sum _{k}\delta _{3k}a_{k3}\end{bmatrix}}\\&=\sum _{i,j,k}\det {\begin{bmatrix}\delta _{1i}&\delta _{1j}&\delta _{1k}\\\delta _{2i}&\delta _{2j}&\delta _{2k}\\\delta _{3i}&\delta _{3j}&\delta _{3k}\end{bmatrix}}a_{i1}a_{j2}a_{k3}\\&=\sum _{i,j,k}\varepsilon _{ijk}a_{i1}a_{j2}a_{k3}\end{aligned}}}