Laplace 展开是行列式计算中的一种重要方法，它通过将行列式展开为子行列式的线性组合，从而简化计算过程。
# 余子式令 A = ( a i j ) A = (a_{ij}) A = ( a i j ) n n n i i i j j j ( n − 1 ) (n-1) ( n − 1 ) 子矩阵 ，记作 A i j A_{ij} A i j
A i j = ( a 11 ⋯ a 1 , j − 1 a 1 , j + 1 ⋯ a 1 n ⋮ ⋮ ⋮ ⋮ a i − 1 , 1 ⋯ a i − 1 , j − 1 a i − 1 , j + 1 ⋯ a i − 1 , n a i + 1 , 1 ⋯ a i + 1 , j − 1 a i + 1 , j + 1 ⋯ a i + 1 , n ⋮ ⋮ ⋮ ⋮ a n 1 ⋯ a n , j − 1 a n , j + 1 ⋯ a n n ) A_{ij} = \begin{pmatrix} a_{11} & \cdots & a_{1,j-1} & a_{1,j+1} & \cdots & a_{1n} \\ \vdots & & \vdots & \vdots & & \vdots \\ a_{i-1,1} & \cdots & a_{i-1,j-1} & a_{i-1,j+1} & \cdots & a_{i-1,n} \\ a_{i+1,1} & \cdots & a_{i+1,j-1} & a_{i+1,j+1} & \cdots & a_{i+1,n} \\ \vdots & & \vdots & \vdots & & \vdots \\ a_{n1} & \cdots & a_{n,j-1} & a_{n,j+1} & \cdots & a_{nn} \end{pmatrix} A i j = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ a 1 1 ⋮ a i − 1 , 1 a i + 1 , 1 ⋮ a n 1 ⋯ ⋯ ⋯ ⋯ a 1 , j − 1 ⋮ a i − 1 , j − 1 a i + 1 , j − 1 ⋮ a n , j − 1 a 1 , j + 1 ⋮ a i − 1 , j + 1 a i + 1 , j + 1 ⋮ a n , j + 1 ⋯ ⋯ ⋯ ⋯ a 1 n ⋮ a i − 1 , n a i + 1 , n ⋮ a n n ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
称该行列式 ∣ A i j ∣ |A_{ij}| ∣ A i j ∣ a i j a_{ij} a i j 余子式 (Minor)「余因子」
行列式可以由余因子写出，这个过程被称为 Laplace 展开「余因子展開」
定理 Laplace 展开定理 A = ( a i j ) A = (a_{ij}) A = ( a i j ) n n n
∣ A ∣ = ∑ j = 1 n a i j ( − 1 ) i + j ∣ A i j ∣ |A| = \sum_{j=1}^n a_{ij} (-1)^{i+j} |A_{ij}| ∣ A ∣ = j = 1 ∑ n a i j ( − 1 ) i + j ∣ A i j ∣
∣ A ∣ = ∑ i = 1 n a i j ( − 1 ) i + j ∣ A i j ∣ |A| = \sum_{i=1}^n a_{ij} (-1)^{i+j} |A_{ij}| ∣ A ∣ = i = 1 ∑ n a i j ( − 1 ) i + j ∣ A i j ∣
证明 对第 i i i
∣ A ∣ = ∑ σ ∈ S n sgn ( σ ) ∏ k = 1 n a k , σ ( k ) = ∑ j = 1 n ∑ σ ∈ S n σ ( i ) = j sgn ( σ ) ∏ k = 1 n a k , σ ( k ) = ∑ j = 1 n a i j ∑ σ ∈ S n σ ( i ) = j sgn ( σ ) ∏ k = 1 k ≠ i n a k , σ ( k ) \begin{aligned} |A| &= \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{k=1}^n a_{k, \sigma(k)} \\ &= \sum_{j=1}^n \sum_{\substack{\sigma \in S_n \\ \sigma(i) = j}} \text{sgn}(\sigma) \prod_{k=1}^n a_{k, \sigma(k)} \\ &= \sum_{j=1}^n a_{ij} \sum_{\substack{\sigma \in S_n \\ \sigma(i) = j}} \text{sgn}(\sigma) \prod_{\substack{k=1 \\ k \neq i}}^n a_{k, \sigma(k)} \end{aligned} ∣ A ∣ = σ ∈ S n ∑ sgn ( σ ) k = 1 ∏ n a k , σ ( k ) = j = 1 ∑ n σ ∈ S n σ ( i ) = j ∑ sgn ( σ ) k = 1 ∏ n a k , σ ( k ) = j = 1 ∑ n a i j σ ∈ S n σ ( i ) = j ∑ sgn ( σ ) k = 1 k = i ∏ n a k , σ ( k )
对于固定的 j j j
ϕ : { σ ∈ S n ∣ σ ( i ) = j } → S n − 1 σ ↦ σ ′ \begin{aligned} \phi: \{\sigma \in S_n \mid \sigma(i) = j\} &\to S_{n-1} \\ \sigma &\mapsto \sigma' \end{aligned} ϕ : { σ ∈ S n ∣ σ ( i ) = j } σ → S n − 1 ↦ σ ′
其中 σ ′ \sigma' σ ′ { 1 , 2 , … , n } ∖ { i } \{1, 2, \dots, n\} \setminus \{i\} { 1 , 2 , … , n } ∖ { i } { 1 , 2 , … , n } ∖ { j } \{1, 2, \dots, n\} \setminus \{j\} { 1 , 2 , … , n } ∖ { j } k ≠ i k \neq i k = i
σ ′ ( k ) = { σ ( k ) , σ ( k ) < j σ ( k ) − 1 , σ ( k ) > j \sigma'(k) = \begin{cases}\sigma(k), & \sigma(k) < j \\ \sigma(k) - 1, & \sigma(k) > j \end{cases} σ ′ ( k ) = { σ ( k ) , σ ( k ) − 1 , σ ( k ) < j σ ( k ) > j
显然 ϕ \phi ϕ σ ′ = ϕ ( σ ) \sigma' = \phi(\sigma) σ ′ = ϕ ( σ )
sgn ( σ ) = ( − 1 ) i + j sgn ( σ ′ ) \text{sgn}(\sigma) = (-1)^{i+j} \text{sgn}(\sigma') sgn ( σ ) = ( − 1 ) i + j sgn ( σ ′ )
因此
∣ A ∣ = ∑ j = 1 n a i j ( − 1 ) i + j ∑ σ ′ ∈ S n − 1 sgn ( σ ′ ) ∏ k = 1 n − 1 a k ′ , σ ′ ( k ′ ) = ∑ j = 1 n a i j ( − 1 ) i + j ∣ A i j ∣ \begin{aligned} |A| &= \sum_{j=1}^n a_{ij} (-1)^{i+j} \sum_{\sigma' \in S_{n-1}} \text{sgn}(\sigma') \prod_{k=1}^{n-1} a_{k', \sigma'(k')} \\ &= \sum_{j=1}^n a_{ij} (-1)^{i+j} |A_{ij}| \end{aligned} ∣ A ∣ = j = 1 ∑ n a i j ( − 1 ) i + j σ ′ ∈ S n − 1 ∑ sgn ( σ ′ ) k = 1 ∏ n − 1 a k ′ , σ ′ ( k ′ ) = j = 1 ∑ n a i j ( − 1 ) i + j ∣ A i j ∣
□ \square □
实际上，这只是对选取的行或者列上的元素进行挨个处理，变成 a i j ∣ A i j ∣ a_{ij} |A_{ij}| a i j ∣ A i j ∣ ( + − + − ⋯ − + − + ⋯ + − + − ⋯ − + − + ⋯ ⋮ ⋮ ⋮ ⋮ ⋱ ) \begin{pmatrix} +& - & + & - & \cdots \\ -& + & - & + & \cdots \\ +& - & + & - & \cdots \\ -& + & - & + & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ + − + − ⋮ − + − + ⋮ + − + − ⋮ − + − + ⋮ ⋯ ⋯ ⋯ ⋯ ⋱ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
在具备 Laplace 展开这样的操作后，求解复杂行列式的方针变为了
利用行基本变换快速降次 或者利用行基本变换尽可能在同一行 / 列中制造出更多的零 选取那一行，进行展开降次 示例
∣ A ∣ = ∣ 2 2 0 2 1 1 0 − 1 3 0 − 2 1 6 14 2 6 ∣ |A| = \begin{vmatrix} 2 & 2 & 0 & 2 \\ 1 & 1 & 0 & -1 \\ 3 & 0 & -2 & 1 \\ 6 & 14 & 2 & 6 \end{vmatrix} ∣ A ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 1 3 6 2 1 0 1 4 0 0 − 2 2 2 − 1 1 6 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
解 → R 1 − 2 R 2 → R 1 ∣ 0 0 0 4 1 1 0 − 1 3 0 − 2 1 6 14 2 6 ∣ → E x p a n s i o n o n R 1 − 4 ∣ 1 1 0 3 0 − 2 6 14 2 ∣ → R 2 − 3 R 1 → R 2 − 4 ∣ 1 1 0 0 − 3 − 2 6 14 2 ∣ → R 3 − 6 R 1 → R 3 − 4 ∣ 1 1 0 0 − 3 − 2 0 8 2 ∣ = − 4 ∣ − 3 − 2 8 2 ∣ = − 4 ( − 6 + 16 ) = − 40 \begin{aligned} &\xrightarrow{R1 - 2R2 \to R1} \begin{vmatrix} 0 & 0 & 0 & 4 \\ 1 & 1 & 0 & -1 \\ 3 & 0 & -2 & 1 \\ 6 & 14 & 2 & 6 \end{vmatrix} \\ &\xrightarrow{Expansion on R1} -4 \begin{vmatrix} 1 & 1 & 0 \\ 3 & 0 & -2 \\ 6 & 14 & 2 \end{vmatrix} \\ &\xrightarrow{R2 - 3R1 \to R2} -4 \begin{vmatrix} 1 & 1 & 0 \\ 0 & -3 & -2 \\ 6 & 14 & 2 \end{vmatrix} \\ &\xrightarrow{R3 - 6R1 \to R3} -4 \begin{vmatrix} 1 & 1 & 0 \\ 0 & -3 & -2 \\ 0 & 8 & 2 \end{vmatrix} \\ &= -4 \begin{vmatrix} -3 & -2 \\ 8 & 2 \end{vmatrix} \\ &= -4 (-6 + 16) \\ &= -40 \end{aligned} R 1 − 2 R 2 → R 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 0 1 3 6 0 1 0 1 4 0 0 − 2 2 4 − 1 1 6 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ E x p a n s i o n o n R 1 − 4 ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 3 6 1 0 1 4 0 − 2 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ R 2 − 3 R 1 → R 2 − 4 ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 0 6 1 − 3 1 4 0 − 2 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ R 3 − 6 R 1 → R 3 − 4 ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 0 0 1 − 3 8 0 − 2 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = − 4 ∣ ∣ ∣ ∣ ∣ − 3 8 − 2 2 ∣ ∣ ∣ ∣ ∣ = − 4 ( − 6 + 1 6 ) = − 4 0
□ \square □
通过 Laplace 展开，还可以求解出一类著名的行列式
示例 Vandermonde 行列式
∣ 1 1 1 ⋯ 1 x 1 x 2 x 3 ⋯ x n x 1 2 x 2 2 x 3 2 ⋯ x n 2 ⋮ ⋮ ⋮ ⋮ x 1 n − 1 x 2 n − 1 x 3 n − 1 ⋯ x n n − 1 ∣ = ∏ 1 ≤ i < j ≤ n ( x j − x i ) \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \end{vmatrix} = \prod_{1 \leq i \lt j \leq n} (x_j - x_i) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 x 1 x 1 2 ⋮ x 1 n − 1 1 x 2 x 2 2 ⋮ x 2 n − 1 1 x 3 x 3 2 ⋮ x 3 n − 1 ⋯ ⋯ ⋯ ⋯ 1 x n x n 2 ⋮ x n n − 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 1 ≤ i < j ≤ n ∏ ( x j − x i )
证明 通过归纳法证明，首先在 n = 2 n=2 n = 2
∣ 1 1 x 1 x 2 ∣ = x 2 − x 1 \begin{vmatrix} 1 & 1 \\ x_1 & x_2 \end{vmatrix} = x_2 - x_1 ∣ ∣ ∣ ∣ ∣ 1 x 1 1 x 2 ∣ ∣ ∣ ∣ ∣ = x 2 − x 1
成立
假设对于 n = k n=k n = k n = k + 1 n=k+1 n = k + 1
∣ 1 1 1 ⋯ 1 x 1 x 2 x 3 ⋯ x k + 1 x 1 2 x 2 2 x 3 2 ⋯ x k + 1 2 ⋮ ⋮ ⋮ ⋮ x 1 k x 2 k x 3 k ⋯ x k + 1 k ∣ → R n − x 1 R ( n − 1 ) → R n ∣ 1 1 1 ⋯ 1 x 1 x 2 x 3 ⋯ x k + 1 x 1 2 x 2 2 x 3 2 ⋯ x k + 1 2 ⋮ ⋮ ⋮ ⋮ x 1 k − 1 x 2 k − 1 x 3 k − 1 ⋯ x k + 1 k − 1 0 x 2 − x 1 x 3 − x 1 ⋯ x k + 1 − x 1 ∣ → R ( n − 1 ) − x 1 R ( n − 2 ) → R ( n − 1 ) ∣ 1 1 1 ⋯ 1 x 1 x 2 x 3 ⋯ x k + 1 x 1 2 x 2 2 x 3 2 ⋯ x k + 1 2 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 x 2 − x 1 x 3 − x 1 ⋯ x k + 1 − x 1 ∣ → R k − x 1 R ( k − 1 ) → R ( k ) … → R 2 − x 1 R 1 → R 2 ∣ 1 1 1 ⋯ 1 0 x 2 − x 1 x 3 − x 1 ⋯ x k + 1 − x 1 0 ( x 2 − x 1 ) ( x 2 + x 1 ) 0 ( x 3 − x 1 ) ( x 3 + x 1 ) ⋯ ( x k + 1 − x 1 ) ( x k + 1 + x 1 ) 0 ⋮ ⋮ ⋮ 0 ( x 2 − x 1 ) ( x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 ) ⋯ ( x k + 1 − x 1 ) ( x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ) ∣ \begin{aligned} \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_{k+1} \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_{k+1}^2 \\ \vdots & \vdots & \vdots & & \vdots \\ x_1^{k} & x_2^{k} & x_3^{k} & \cdots & x_{k+1}^{k} \end{vmatrix} &\xrightarrow{Rn - x_1 R(n-1) \to Rn} \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_{k+1} \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_{k+1}^2 \\ \vdots & \vdots & \vdots & & \vdots \\ x_1^{k-1} & x_2^{k-1} & x_3^{k-1} & \cdots & x_{k+1}^{k-1} \\ 0 & x_2 - x_1 & x_3 - x_1 & \cdots & x_{k+1} - x_1 \end{vmatrix} \\ &\xrightarrow{R(n-1) - x_1 R(n-2) \to R(n-1)} \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_{k+1} \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_{k+1}^2 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & x_2 - x_1 & x_3 - x_1 & \cdots & x_{k+1} - x_1 \end{vmatrix} \\ &\xrightarrow{Rk - x_1 R(k-1) \to R(k)} \ldots \\[30pt] &\xrightarrow{R2 - x_1 R1 \to R2} \begin{vmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & x_2 - x_1 & x_3 - x_1 & \cdots & x_{k+1} - x_1 \\ 0 & (x_2 - x_1)(x_2 + x_1) \\ 0 & (x_3 - x_1)(x_3 + x_1) & \cdots & (x_{k+1} - x_1)(x_{k+1} + x_1) \\ 0 & \vdots & \vdots & & \vdots \\ 0 & (x_2 - x_1)(x_2^{k-1} + x_2^{k-2} x_1 + \cdots + x_1^{k-1}) & \cdots & (x_{k+1} - x_1)(x_{k+1}^{k-1} + x_{k+1}^{k-2} x_1 + \cdots + x_1^{k-1}) \end{vmatrix} \end{aligned} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 x 1 x 1 2 ⋮ x 1 k 1 x 2 x 2 2 ⋮ x 2 k 1 x 3 x 3 2 ⋮ x 3 k ⋯ ⋯ ⋯ ⋯ 1 x k + 1 x k + 1 2 ⋮ x k + 1 k ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ R n − x 1 R ( n − 1 ) → R n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 x 1 x 1 2 ⋮ x 1 k − 1 0 1 x 2 x 2 2 ⋮ x 2 k − 1 x 2 − x 1 1 x 3 x 3 2 ⋮ x 3 k − 1 x 3 − x 1 ⋯ ⋯ ⋯ ⋯ ⋯ 1 x k + 1 x k + 1 2 ⋮ x k + 1 k − 1 x k + 1 − x 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ R ( n − 1 ) − x 1 R ( n − 2 ) → R ( n − 1 ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 x 1 x 1 2 ⋮ 0 0 1 x 2 x 2 2 ⋮ 0 x 2 − x 1 1 x 3 x 3 2 ⋮ 0 x 3 − x 1 ⋯ ⋯ ⋯ ⋯ ⋯ 1 x k + 1 x k + 1 2 ⋮ 0 x k + 1 − x 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ R k − x 1 R ( k − 1 ) → R ( k ) … R 2 − x 1 R 1 → R 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 0 0 0 0 0 1 x 2 − x 1 ( x 2 − x 1 ) ( x 2 + x 1 ) ( x 3 − x 1 ) ( x 3 + x 1 ) ⋮ ( x 2 − x 1 ) ( x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 ) 1 x 3 − x 1 ⋯ ⋮ ⋯ ⋯ ⋯ ( x k + 1 − x 1 ) ( x k + 1 + x 1 ) ( x k + 1 − x 1 ) ( x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ) 1 x k + 1 − x 1 ⋮ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
对其第一列进行 Laplace 展开
→ Expansion on C1 ∣ x 2 − x 1 x 3 − x 1 ⋯ x k + 1 − x 1 ( x 2 − x 1 ) ( x 2 + x 1 ) ( x 3 − x 1 ) ( x 3 + x 1 ) ⋯ ( x k + 1 − x 1 ) ( x k + 1 + x 1 ) ⋮ ⋮ ⋮ ( x 2 − x 1 ) ( x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 ) ( x 3 − x 1 ) ( x 3 k − 1 + x 3 k − 2 x 1 + ⋯ + x 1 k − 1 ) ⋯ ( x k + 1 − x 1 ) ( x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ) ∣ = ∏ j = 2 k + 1 ( x j − x 1 ) ∣ 1 1 ⋯ 1 x 2 + x 1 x 3 + x 1 ⋯ x k + 1 + x 1 ⋮ ⋮ ⋮ x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 x 3 k − 1 + x 3 k − 2 x 1 + ⋯ + x 1 k − 1 ⋯ x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ∣ \begin{aligned} &\xrightarrow{\text{Expansion on C1}} \begin{vmatrix} x_2 - x_1 & x_3 - x_1 & \cdots & x_{k+1} - x_1 \\ (x_2 - x_1)(x_2 + x_1) & (x_3 - x_1)(x_3 + x_1) & \cdots & (x_{k+1} - x_1)(x_{k+1} + x_1) \\ \vdots & \vdots & & \vdots \\ (x_2 - x_1)(x_2^{k-1} + x_2^{k-2} x_1 + \cdots + x_1^{k-1}) & (x_3 - x_1)(x_3^{k-1} + x_3^{k-2} x_1 + \cdots + x_1^{k-1}) & \cdots & (x_{k+1} - x_1)(x_{k+1}^{k-1} + x_{k+1}^{k-2} x_1 + \cdots + x_1^{k-1}) \end{vmatrix} \\ &= \prod_{j=2}^{k+1} (x_j - x_1) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_2 + x_1 & x_3 + x_1 & \cdots & x_{k+1} + x_1 \\ \vdots & \vdots & & \vdots \\ x_2^{k-1} + x_2^{k-2} x_1 + \cdots + x_1^{k-1} & x_3^{k-1} + x_3^{k-2} x_1 + \cdots + x_1^{k-1} & \cdots & x_{k+1}^{k-1} + x_{k+1}^{k-2} x_1 + \cdots + x_1^{k-1} \end{vmatrix} \end{aligned} Expansion on C1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ x 2 − x 1 ( x 2 − x 1 ) ( x 2 + x 1 ) ⋮ ( x 2 − x 1 ) ( x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 ) x 3 − x 1 ( x 3 − x 1 ) ( x 3 + x 1 ) ⋮ ( x 3 − x 1 ) ( x 3 k − 1 + x 3 k − 2 x 1 + ⋯ + x 1 k − 1 ) ⋯ ⋯ ⋯ x k + 1 − x 1 ( x k + 1 − x 1 ) ( x k + 1 + x 1 ) ⋮ ( x k + 1 − x 1 ) ( x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = j = 2 ∏ k + 1 ( x j − x 1 ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 x 2 + x 1 ⋮ x 2 k − 1 + x 2 k − 2 x 1 + ⋯ + x 1 k − 1 1 x 3 + x 1 ⋮ x 3 k − 1 + x 3 k − 2 x 1 + ⋯ + x 1 k − 1 ⋯ ⋯ ⋯ 1 x k + 1 + x 1 ⋮ x k + 1 k − 1 + x k + 1 k − 2 x 1 + ⋯ + x 1 k − 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
由归纳假设可知
= ∏ j = 2 k + 1 ( x j − x 1 ) ∏ 2 ≤ i < j ≤ k + 1 ( x j − x i ) = ∏ 1 ≤ i < j ≤ k + 1 ( x j − x i ) \begin{aligned} &= \prod_{j=2}^{k+1} (x_j - x_1) \prod_{2 \leq i \lt j \leq k+1} (x_j - x_i) \\ &= \prod_{1 \leq i \lt j \leq k+1} (x_j - x_i) \end{aligned} = j = 2 ∏ k + 1 ( x j − x 1 ) 2 ≤ i < j ≤ k + 1 ∏ ( x j − x i ) = 1 ≤ i < j ≤ k + 1 ∏ ( x j − x i )
□ \square □
# 伴随矩阵对于 n n n A = ( a i j ) A = (a_{ij}) A = ( a i j ) ( i , j ) (i, j) ( i , j ) a j i a_{ji} a j i
adj ( A ) = ( ( − 1 ) i + j ∣ A j i ∣ ) = ( ∣ A 11 ∣ − ∣ A 21 ∣ ∣ A 31 ∣ ⋯ ( − 1 ) 1 + n ∣ A n 1 ∣ − ∣ A 12 ∣ ∣ A 22 ∣ − ∣ A 32 ∣ ⋯ ( − 1 ) 2 + n ∣ A n 2 ∣ ∣ A 13 ∣ − ∣ A 23 ∣ ∣ A 33 ∣ ⋯ ( − 1 ) 3 + n ∣ A n 3 ∣ ⋮ ⋮ ⋮ ⋮ ( − 1 ) n + 1 ∣ A 1 n ∣ ( − 1 ) n + 2 ∣ A 2 n ∣ ( − 1 ) n + 3 ∣ A 3 n ∣ ⋯ ∣ A n n ∣ ) \text{adj}(A) = \left((-1)^{i+j} |A_{ji}|\right) = \begin{pmatrix} |A_{11}| & -|A_{21}| & |A_{31}| & \cdots & (-1)^{1+n} |A_{n1}| \\ -|A_{12}| & |A_{22}| & -|A_{32}| & \cdots & (-1)^{2+n} |A_{n2}| \\ |A_{13}| & -|A_{23}| & |A_{33}| & \cdots & (-1)^{3+n} |A_{n3}| \\ \vdots & \vdots & \vdots & & \vdots \\ (-1)^{n+1} |A_{1n}| & (-1)^{n+2} |A_{2n}| & (-1)^{n+3} |A_{3n}| & \cdots & |A_{nn}| \end{pmatrix} adj ( A ) = ( ( − 1 ) i + j ∣ A j i ∣ ) = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ ∣ A 1 1 ∣ − ∣ A 1 2 ∣ ∣ A 1 3 ∣ ⋮ ( − 1 ) n + 1 ∣ A 1 n ∣ − ∣ A 2 1 ∣ ∣ A 2 2 ∣ − ∣ A 2 3 ∣ ⋮ ( − 1 ) n + 2 ∣ A 2 n ∣ ∣ A 3 1 ∣ − ∣ A 3 2 ∣ ∣ A 3 3 ∣ ⋮ ( − 1 ) n + 3 ∣ A 3 n ∣ ⋯ ⋯ ⋯ ⋯ ( − 1 ) 1 + n ∣ A n 1 ∣ ( − 1 ) 2 + n ∣ A n 2 ∣ ( − 1 ) 3 + n ∣ A n 3 ∣ ⋮ ∣ A n n ∣ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
称该矩阵为 A A A 伴随矩阵 (Adjoint Matrix)「余因子行列」
伴随矩阵的一个重要作用是求解逆矩阵
命题
A ⋅ adj ( A ) = adj ( A ) ⋅ A = ∣ A ∣ E n A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| E_n A ⋅ adj ( A ) = adj ( A ) ⋅ A = ∣ A ∣ E n
证明 考虑乘积 A ⋅ adj ( A ) A \cdot \text{adj}(A) A ⋅ adj ( A ) ( i , j ) (i, j) ( i , j )
( A ⋅ adj ( A ) ) i j = ∑ k = 1 n a i k ⋅ ( − 1 ) k + j ∣ A j k ∣ (A \cdot \text{adj}(A))_{ij} = \sum_{k=1}^n a_{ik} \cdot (-1)^{k+j} |A_{jk}| ( A ⋅ adj ( A ) ) i j = k = 1 ∑ n a i k ⋅ ( − 1 ) k + j ∣ A j k ∣
当 i = j i = j i = j
( A ⋅ adj ( A ) ) i i = ∑ k = 1 n a i k ⋅ ( − 1 ) k + i ∣ A i k ∣ = ∣ A ∣ (A \cdot \text{adj}(A))_{ii} = \sum_{k=1}^n a_{ik} \cdot (-1)^{k+i} |A_{ik}| = |A| ( A ⋅ adj ( A ) ) i i = k = 1 ∑ n a i k ⋅ ( − 1 ) k + i ∣ A i k ∣ = ∣ A ∣
这正是对第 i i i A ⋅ adj ( A ) A \cdot \text{adj}(A) A ⋅ adj ( A ) ∣ A ∣ |A| ∣ A ∣
当 i ≠ j i \neq j i = j
B = ( a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a j 1 a j 2 ⋯ a j n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ) B = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{j1} & a_{j2} & \cdots & a_{jn} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} B = ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎛ a 1 1 ⋮ a j 1 ⋮ a i 1 ⋮ a n 1 a 1 2 ⋮ a j 2 ⋮ a i 2 ⋮ a n 2 ⋯ ⋯ ⋯ ⋯ a 1 n ⋮ a j n ⋮ a i n ⋮ a n n ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎞
即将 A A A i i i j j j B B B j j j
∣ B ∣ = ∑ k = 1 n a i k ⋅ ( − 1 ) k + j ∣ A j k ∣ |B| = \sum_{k=1}^n a_{ik} \cdot (-1)^{k+j} |A_{jk}| ∣ B ∣ = k = 1 ∑ n a i k ⋅ ( − 1 ) k + j ∣ A j k ∣
但是由于 B B B i i i j j j ∣ B ∣ = − ∣ A ∣ |B| = -|A| ∣ B ∣ = − ∣ A ∣ ∣ A ∣ |A| ∣ A ∣
( A ⋅ adj ( A ) ) i j = 0 (A \cdot \text{adj}(A))_{ij} = 0 ( A ⋅ adj ( A ) ) i j = 0
这意味着 A ⋅ adj ( A ) A \cdot \text{adj}(A) A ⋅ adj ( A ) 0 0 0 A ⋅ adj ( A ) = ∣ A ∣ E n A \cdot \text{adj}(A) = |A| E_n A ⋅ adj ( A ) = ∣ A ∣ E n adj ( A ) ⋅ A = ∣ A ∣ E n \text{adj}(A) \cdot A = |A| E_n adj ( A ) ⋅ A = ∣ A ∣ E n □ \square □
所以，如果 A A A
A ( 1 ∣ A ∣ adj ( A ) ) = 1 ∣ A ∣ adj ( A ) A = E n A \left(\frac{1}{|A|} \text{adj}(A)\right) = \frac{1}{|A|} \text{adj}(A) A = E_n A ( ∣ A ∣ 1 adj ( A ) ) = ∣ A ∣ 1 adj ( A ) A = E n
即
A − 1 = 1 ∣ A ∣ adj ( A ) A^{-1} = \frac{1}{|A|} \text{adj}(A) A − 1 = ∣ A ∣ 1 adj ( A )
