（邱维声）高等代数课程笔记：行列式按k行（列）展开
行列式按 k 行（列）展开
k 阶子式与其余子式：对于 n n n 级矩阵 A = ( a i j ) A=(a_{ij}) A=(aij)，任取其 k k k 行：第 i 1 , i 2 , ⋯ , i k i_{1},i_{2},\cdots,i_{k} i1,i2,⋯,ik 行，其中 i 1 < i 2 < ⋯ < i k i_{1}
A ( i 1 , i 2 , ⋯ , i k j 1 , j 2 , ⋯ , j k ) (1) A\left(\begin{matrix} i_{1},i_{2},\cdots,i_{k}\\ j_{1},j_{2},\cdots,j_{k} \end{matrix}\right) \tag{1} A(i1,i2,⋯,ikj1,j2,⋯,jk)(1)
\quad 若划去上述 k k k 行 k k k 列，则剩下元素按照原来的排法可以形成一个 n − k n-k n−k 阶行列式，称为 ( 1 ) (1) (1) 的余子式。
\quad 令
{ i 1 ′ , i 2 ′ , ⋯ , i n − k ′ } = { 1 , 2 , ⋯ , n } \ { i 1 , i 2 , ⋯ , i k } \{i_{1}',i_{2}',\cdots,i_{n-k}'\} = \{1,2,\cdots,n\} \backslash \{i_{1},i_{2},\cdots,i_{k}\} {i1′,i2′,⋯,in−k′}={1,2,⋯,n}\{i1,i2,⋯,ik}
{ j 1 ′ , j 2 ′ , ⋯ , j n − k ′ } = { 1 , 2 , ⋯ , n } \ { j 1 , j 2 , ⋯ , j k } \{j_{1}',j_{2}',\cdots,j_{n-k}'\} = \{1,2,\cdots,n\} \backslash \{j_{1},j_{2},\cdots,j_{k}\} {j1′,j2′,⋯,jn−k′}={1,2,⋯,n}\{j1,j2,⋯,jk}
并且约定： i 1 ′ < i 2 ′ < ⋯ < i n − k ′ i_{1}'
A ( i 1 ′ , i 2 ′ , ⋯ , i n − k ′ j 1 ′ , j 2 ′ , ⋯ , j n − k ′ ) (2) A\left(\begin{matrix} i_{1}',i_{2}',\cdots,i_{n-k}'\\ j_{1}',j_{2}',\cdots,j_{n-k}' \end{matrix}\right)\tag{2} A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)(2)
\quad 另外，称
( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯ , i n − k ′ j 1 ′ , j 2 ′ , ⋯ , j n − k ′ ) (-1)^{(i_{1}+i_{2}+\cdots +i_{k}) + (j_{1} + j_{2} +\cdots + j_{k})} \cdot A\left(\begin{matrix} i_{1}',i_{2}',\cdots,i_{n-k}'\\ j_{1}',j_{2}',\cdots,j_{n-k}'\end{matrix}\right) (−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)
为子式 ( 1 ) (1) (1) 的代数余子式。
\quad 类比 上一节 中 n n n 阶行列式按一行（列）展开的理论，容易猜测 定理 1 成立。
定理 1. Laplace 定理：设 A = ( a i j ) A=(a_{ij}) A=(aij) 是一个 n n n 级矩阵，取其第 i 1 , i 2 , ⋯ , i k i_{1},i_{2},\cdots,i_{k} i1,i2,⋯,ik 行（ i 1 < i 2 < ⋯ < i k i_{1}
∣ A ∣ = ∑ 1 ≤ j 1 < j 2 < ⋯ < j k ≤ n A ( i 1 , i 2 , ⋯ , i k j 1 , j 2 , ⋯ , j k ) ⋅ ( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯ , i n − k ′ j 1 ′ , j 2 ′ , ⋯ , j n − k ′ ) |A| = \sum_{1\le j_{1}
证明：此处是在课上邱老师给出的新证明。
\quad 对于指定的第 i 1 , i 2 , ⋯ , i k i_{1},i_{2},\cdots,i_{k} i1,i2,⋯,ik 行，按照 j 1 < j 2 < ⋯ < j k j_{1}
- 对于任取的第 j 1 , j 2 , ⋯ , j k j_{1},j_{2},\cdots,j_{k} j1,j2,⋯,jk 列，令 u 1 u 2 ⋯ u k u_{1}u_{2}\cdots u_{k} u1u2⋯uk 是其一个 k k k 元排列；
- 对于剩下的第 j 1 ′ < j 2 ′ < ⋯ < j n − k ′ j_{1}'
- 这样一来， ∀ u ∈ { u 1 , u 2 , ⋯ , u k } , ∀ v ∈ { v 1 , v 2 , ⋯ , v n − k } : u < v \forall ~ u \in \{u_{1},u_{2},\cdots,u_{k}\},\forall ~ v \in \{v_{1},v_{2},\cdots,v_{n-k}\}:u
∀ u∈{u1,u2,⋯,uk},∀ v∈{v1,v2,⋯,vn−k}:u<v .
\quad 于是，按照行列式的定义，
∣ A ∣ = ∑ u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ∑ u 1 u 2 ⋯ u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k \begin{aligned} |A| &= \sum_{u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k}}(-1)^{\tau(i_{1}i_{2}\cdots i_{k}i_{1}'i_{2}'\cdots i_{n-k}')+\tau(u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k})} a_{i_{1}u_{1}}a_{i_{2}u_{2}}\cdots a_{i_{k}u_{k}}a_{i_{1}'v_{1}}a_{i_{2}'v_{2}}\cdots a_{i_{n-k}'v_{n-k}} \\ &= \sum_{1\le j_{1}
\quad 要注意到，
a. 对于 n n n 元排列 i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ i_{1}i_{2}\cdots i_{k} i_{1}'i_{2}'\cdots i_{n-k}' i1i2⋯iki1′i2′⋯in−k′，由于比 i 1 i_{1} i1 小的数有 i 1 − 1 i_{1}-1 i1−1 个，比 i 2 i_{2} i2 小的数有 i 2 − 1 i_{2}-1 i2−1 个，再除去 i 1 i_{1} i1，有 i 2 − 2 i_{2}-2 i2−2 个 ……，比 i k i_{k} ik 小的数有 i k − k i_{k}-k ik−k 个，于是
( − 1 ) i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ = ( − 1 ) i 1 + i 2 + ⋯ + i k − k ( k + 1 ) 2 (-1)^{i_{1}i_{2}\cdots i_{k} i_{1}'i_{2}'\cdots i_{n-k}'} = (-1)^{i_{1}+i_{2}+\cdots +i_{k} - \frac{k(k+1)}{2}} (−1)i1i2⋯iki1′i2′⋯in−k′=(−1)i1+i2+⋯+ik−2k(k+1)
b. 假设经过 S S S 次对换，排列 u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k} u1u2⋯ukv1v2⋯vn−k 化为排列 j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k j_{1}j_{2}\cdots j_{k} v_{1}v_{2}\cdots v_{n-k} j1j2⋯jkv1v2⋯vn−k. 因为 j 1 < j 2 < ⋯ < j k j_{1}
( − 1 ) τ ( u 1 u 2 ⋯ u k ) = ( − 1 ) S (-1)^{\tau(u_{1}u_{2}\cdots u_{k})} = (-1)^{S} (−1)τ(u1u2⋯uk)=(−1)S
\quad 于是
( − 1 ) τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) = ( − 1 ) S ⋅ ( − 1 ) τ ( j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k ) = ( − 1 ) τ ( u 1 u 2 ⋯ u k ) ⋅ ( − 1 ) τ ( j 1 j 2 ⋯ j k v 1 v 2 ⋯ v n − k ) = ( − 1 ) j 1 + j 2 + ⋯ + j k − k ( k + 1 ) 2 ⋅ ( − 1 ) τ ( u 1 u 2 ⋯ u k ) + τ ( v 1 v 2 ⋯ v n − k ) \begin{aligned} (-1)^{\tau(u_{1}u_{2}\cdots u_{k}v_{1}v_{2}\cdots v_{n-k})} &= (-1)^{S} \cdot (-1)^{\tau(j_{1}j_{2}\cdots j_{k}v_{1}v_{2}\cdots v_{n-k})}\\ &= (-1)^{\tau(u_{1}u_{2}\cdots u_{k})}\cdot (-1)^{\tau(j_{1}j_{2}\cdots j_{k}v_{1}v_{2}\cdots v_{n-k})}\\ &= (-1)^{j_{1}+j_{2}+\cdots + j_{k} - \frac{k(k+1)}{2}} \cdot (-1)^{\tau(u_{1}u_{2}\cdots u_{k})+ \tau(v_{1}v_{2}\cdots v_{n-k})} \end{aligned} (−1)τ(u1u2⋯ukv1v2⋯vn−k)=(−1)S⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)τ(u1u2⋯uk)⋅(−1)τ(j1j2⋯jkv1v2⋯vn−k)=(−1)j1+j2+⋯+jk−2k(k+1)⋅(−1)τ(u1u2⋯uk)+τ(v1v2⋯vn−k)
\quad 从而，
∣ A ∣ = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ∑ u 1 u 2 ⋯ u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( i 1 i 2 ⋯ i k i 1 ′ i 2 ′ ⋯ i n − k ′ ) + τ ( u 1 u 2 ⋯ u k v 1 v 2 ⋯ v n − k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j n ≤ n ( − 1 ) τ ( i 1 + i 2 + ⋯ + i k ) + τ ( j 1 + j 2 + ⋯ + j k ) ∑ u 1 u 2 ⋯ u k ( − 1 ) τ ( u 1 u 2 ⋯ u k ) a i 1 u 1 a i 2 u 2 ⋯ a i k u k ∑ v 1 v 2 ⋯ v n − k ( − 1 ) τ ( v 1 v 2 ⋯ v n − k ) a i 1 ′ v 1 a i 2 ′ v 2 ⋯ a i n − k ′ v n − k = ∑ 1 ≤ j 1 < j 2 < ⋯ < j k ≤ n ( − 1 ) ( i 1 + i 2 + ⋯ + i k ) + ( j 1 + j 2 + ⋯ + j k ) ⋅ A ( i 1 , i 2 , ⋯ , i k j 1 , j 2 , ⋯ , j k ) ⋅ A ( i 1 ′ , i 2 ′ , ⋯ , i n − k ′ j 1 ′ , j 2 ′ , ⋯ , j n − k ′ ) \begin{aligned} |A| &= \sum_{1\le j_{1}
#
\quad 由行列式的行与列的对称性，不难将结论推广到列的情形。
\quad 最后，介绍 Laplace 定理 的一个重要应用。
例题 1：求解下面的行列式。
∣ A ∣ = ∣ a 11 ⋯ a 1 k 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ a k 1 ⋯ a k k 0 ⋯ 0 c 11 ⋯ c 1 k b 11 ⋯ b 1 t ⋮ ⋮ ⋮ ⋮ c t 1 ⋯ c t k b t 1 ⋯ b t t ∣ |A| = \left|\begin{matrix} a_{11} &\cdots &a_{1k} &0 &\cdots &0\\ \vdots & &\vdots &\vdots & &\vdots\\ a_{k1} &\cdots &a_{kk} &0 &\cdots &0\\ c_{11} &\cdots &c_{1k} &b_{11} &\cdots &b_{1t}\\ \vdots & &\vdots &\vdots & &\vdots\\ c_{t1} &\cdots &c_{tk} &b_{t1} &\cdots &b_{tt} \end{matrix}\right| ∣A∣= a11⋮ak1c11⋮ct1⋯⋯⋯⋯a1k⋮akkc1k⋮ctk0⋮0b11⋮bt1⋯⋯⋯⋯0⋮0b1t⋮btt
解：
\quad 由 Laplace 定理，按前 k k k 行展开，则有：
∣ A ∣ = ∣ a 11 ⋯ a 1 k ⋮ ⋮ a k 1 ⋯ a k k ∣ ⋅ ∣ b 11 ⋯ b 1 t ⋮ ⋮ b t 1 ⋯ b t t ∣ |A| = \left|\begin{matrix}a_{11} &\cdots &a_{1k}\\ \vdots & &\vdots \\ a_{k1} &\cdots &a_{kk}\end{matrix}\right| \cdot \left|\begin{matrix}b_{11} &\cdots &b_{1t}\\ \vdots & &\vdots \\ b_{t1} &\cdots &b_{tt}\end{matrix}\right| ∣A∣= a11⋮ak1⋯⋯a1k⋮akk ⋅ b11⋮bt1⋯⋯b1t⋮btt
#
参考：
- 邱维声. 高等代数课程.