余因子展開
det A = | a 1 , 1 ⋯ a 1 , j ⋯ a 1 , n ⋮ ⋮ ⋮ a i , 1 ⋯ a i , j ⋯ a i , n ⋮ ⋮ ⋮ a n , 1 ⋯ a n , j ⋯ a n , n | = ∑ i = 1 n a i , j | a 1 , 1 ⋯ 0 ⋯ a 1 , n ⋮ ⋮ ⋮ a i , 1 ⋯ 1 ⋯ a i , n ⋮ ⋮ ⋮ a n , 1 ⋯ 0 ⋯ a n , n | = ∑ i = 1 n a i , j ⋅ ( − 1 ) ( i − 1 ) + ( j − 1 ) | 1 0 ⋯ 0 ̆ ⋯ 0 0 a 1 , 1 ⋯ a ̆ 1 , j ⋯ a 1 , n ⋮ ⋮ ⋮ ⋮ 0 ̆ a ̆ i , 1 ⋯ a ̆ i , j ⋯ a ̆ i , n ⋮ ⋮ ⋮ ⋮ 0 a n , 1 ⋯ a ̆ n , j ⋯ a n , n | = ∑ i = 1 n a i , j ⋅ ( − 1 ) i + j | a 1 , 1 ⋯ a ̆ 1 , j ⋯ a 1 , n ⋮ ⋮ ⋮ a ̆ i , 1 ⋯ a ̆ i , j ⋯ a ̆ i , n ⋮ ⋮ ⋮ a n , 1 ⋯ a ̆ n , j ⋯ a n , n | = ∑ i = 1 n a i , j a ~ i , j ◼ {\displaystyle {\begin{aligned}\det A&={\begin{vmatrix}a_{1,1}&\cdots &a_{1,j}&\cdots &a_{1,n}\\\vdots &&\vdots &&\vdots \\a_{i,1}&\cdots &a_{i,j}&\cdots &a_{i,n}\\\vdots &&\vdots &&\vdots \\a_{n,1}&\cdots &a_{n,j}&\cdots &a_{n,n}\\\end{vmatrix}}\\&\\&=\textstyle \sum \limits _{i=1}^{n}\,a_{i,j}\,{\begin{vmatrix}a_{1,1}&\cdots &0&\cdots &a_{1,n}\\\vdots &&\vdots &&\vdots \\a_{i,1}&\cdots &1&\cdots &a_{i,n}\\\vdots &&\vdots &&\vdots \\a_{n,1}&\cdots &0&\cdots &a_{n,n}\\\end{vmatrix}}\\&\\&=\textstyle \sum \limits _{i=1}^{n}\,a_{i,j}\,\cdot (-1)^{(i-1)+(j-1)}\,{\begin{vmatrix}\;1&0&\cdots &{\breve {0}}&\cdots &0\\\;0&a_{1,1}&\cdots &{\breve {a}}_{1,j}&\cdots &a_{1,n}\\\;\vdots &\vdots &&\vdots &&\vdots \\\;{\breve {0}}&{\breve {a}}_{i,1}&\cdots &{\breve {a}}_{i,j}&\cdots &{\breve {a}}_{i,n}\\\;\vdots &\vdots &&\vdots &&\vdots \\\;0&a_{n,1}&\cdots &{\breve {a}}_{n,j}&\cdots &a_{n,n}\\\end{vmatrix}}\\&\\&=\textstyle \sum \limits _{i=1}^{n}\,a_{i,j}\,\cdot (-1)^{i+j}\,{\begin{vmatrix}a_{1,1}&\cdots &{\breve {a}}_{1,j}&\cdots &a_{1,n}\\\vdots &&\vdots &&\vdots \\{\breve {a}}_{i,1}&\cdots &{\breve {a}}_{i,j}&\cdots &{\breve {a}}_{i,n}\\\vdots &&\vdots &&\vdots \\a_{n,1}&\cdots &{\breve {a}}_{n,j}&\cdots &a_{n,n}\\\end{vmatrix}}\\&=\textstyle \sum \limits _{i=1}^{n}a_{i,j}\,{\widetilde {a}}_{i,j}\quad \blacksquare \end{aligned}}}
τ ′ = [ 1 ⋯ i − 1 i ⋯ n − 1 n ( ← ) j σ ( 1 ) ⋯ ( ← ) j σ ( i − 1 ) ( ← ) j σ ( i + 1 ) ⋯ ( ← ) j σ ( n ) n ] {\displaystyle \tau '={\begin{bmatrix}1&\cdots &i-1&i&\cdots &n-1&n\\(\leftarrow )_{j}\sigma (1)&\cdots &(\leftarrow )_{j}\sigma (i-1)&(\leftarrow )_{j}\sigma (i+1)&\cdots &(\leftarrow )_{j}\sigma (n)&n\end{bmatrix}}}
と表せる。このとき、先に (←)i(これは巡回置換 (n, n − 1, ..., i + 1, i) のことである)を施してから τ′ を施す置換 τ′(←)i も、σ を施してから (←)j を施す置換 (←)jσ も、どちらも次の置換になる:
[ 1 ⋯ i − 1 i i + 1 ⋯ n ( ← ) j σ ( 1 ) ⋯ ( ← ) j σ ( i − 1 ) n ( ← ) j σ ( i + 1 ) ⋯ ( ← ) j σ ( n ) ] {\displaystyle {\begin{bmatrix}1&\cdots &i-1&i&i+1&\cdots &n\\(\leftarrow )_{j}\sigma (1)&\cdots &(\leftarrow )_{j}\sigma (i-1)&n&(\leftarrow )_{j}\sigma (i+1)&\cdots &(\leftarrow )_{j}\sigma (n)\end{bmatrix}}}