行列式按 k 行（列）展开

k 阶子式与其余子式：对于 $n$ 级矩阵 A=(aij)A=(a_{ij})A=(aij​)，任取其 $k$ 行：第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1​,i2​,⋯,ik​ 行，其中 i1<i2<⋯<iki_{1}<i_{2}<\\cdots<i_{k}i1​<i2​<⋯<ik​；再任取其 $k$ 列：第 j1,j2,⋯,jkj_{1},j_{2},\\cdots,j_{k}j1​,j2​,⋯,jk​ 列，其中 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots<j_{k}j1​<j2​<⋯<jk​. 则这 $k$ 行 $k$ 列元素按原来的排法可以构成一个 $k$ 阶行列式，称为 $A$ 的一个 $k$ 阶子式，记作：

A(i1,i2,⋯,ikj1,j2,⋯,jk)(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​,⋯,ik​j1​,j2​,⋯,jk​​)(1)

\\quad 若划去上述 $k$ 行 $k$ 列，则剩下元素按照原来的排法可以形成一个 $n-k$ 阶行列式，称为 $(1)$ 的余子式。

\\quad 令

{i1′,i2′,⋯,in−k′}={1,2,⋯,n}\\{i1,i2,⋯,ik}\\{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​}

{j1′,j2′,⋯,jn−k′}={1,2,⋯,n}\\{j1,j2,⋯,jk}\\{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：i1′<i2′<⋯<in−k′i_{1}'<i_{2}'<\\cdots <i_{n-k}'i1′​<i2′​<⋯<in−k′​，j1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots <j_{n-k}'j1′​<j2′​<⋯<jn−k′​. 则 $(1)$ 的余子式也是 $A$ 的一个 $n-k$ 阶子式，记作：

A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−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)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−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 $(1)$ 的代数余子式。

\\quad 类比 上一节 中 $n$ 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=(aij)A=(a_{ij})A=(aij​) 是一个 $n$ 级矩阵，取其第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1​,i2​,⋯,ik​ 行（i1<i2<⋯<iki_{1}<i_{2}<\\cdots <i_{k}i1​<i2​<⋯<ik​），则 $|A|$ 等于这 $k$ 行形成的所有 $k$ 阶子式与其代数余子式的乘积之和。即

∣A∣=∑1≤j1<j2<⋯<jk≤nA(i1,i2,⋯,ikj1,j2,⋯,jk)⋅(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)|A| = \\sum_{1\\le j_{1}<j_{2}<\\cdots<j_{k}\\le n} A\\left(\\begin{matrix}i_{1},i_{2},\\cdots,i_{k}\\\\ j_{1},j_{2},\\cdots,j_{k}\\end{matrix}\\right) \\cdot (-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) ∣A∣=1≤j1​<j2​<⋯<jk​≤n∑​A(i1​,i2​,⋯,ik​j1​,j2​,⋯,jk​​)⋅(−1)(i1​+i2​+⋯+ik​)+(j1​+j2​+⋯+jk​)⋅A(i1′​,i2′​,⋯,in−k′​j1′​,j2′​,⋯,jn−k′​​)

证明：此处是在课上邱老师给出的新证明。

\\quad 对于指定的第 i1,i2,⋯,iki_{1},i_{2},\\cdots,i_{k}i1​,i2​,⋯,ik​ 行，按照 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots <j_{k}j1​<j2​<⋯<jk​ 的要求任取列，从而得到 $k$ 阶子式。对 $k$ 阶子式可以进行“分组”：

• 对于任取的第 j1,j2,⋯,jkj_{1},j_{2},\\cdots,j_{k}j1​,j2​,⋯,jk​ 列，令 u1u2⋯uku_{1}u_{2}\\cdots u_{k}u1​u2​⋯uk​ 是其一个 $k$ 元排列；
• 对于剩下的第 j1′<j2′<⋯<jn−k′j_{1}'<j_{2}'<\\cdots<j_{n-k}'j1′​<j2′​<⋯<jn−k′​ 列，令 v1v2⋯vn−kv_{1}v_{2}\\cdots v_{n-k}v1​v2​⋯vn−k​ 是其一个 $n-k$ 元排列。
• 这样一来，∀u∈{u1,u2,⋯,uk},∀v∈{v1,v2,⋯,vn−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<v∀ u∈{u1​,u2​,⋯,uk​},∀ v∈{v1​,v2​,⋯,vn−k​}:u<v.

\\quad 于是，按照行列式的定义，

∣A∣=∑u1u2⋯ukv1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−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} <j_{2}<\\cdots<j_{n}\\le n}\\sum_{u_{1}u_{2}\\cdots u_{k}}\\sum_{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}} \\end{aligned} ∣A∣​=u1​u2​⋯uk​v1​v2​⋯vn−k​∑​(−1)τ(i1​i2​⋯ik​i1′​i2′​⋯in−k′​)+τ(u1​u2​⋯uk​v1​v2​⋯vn−k​)ai1​u1​​ai2​u2​​⋯aik​uk​​ai1′​v1​​ai2′​v2​​⋯ain−k′​vn−k​​=1≤j1​<j2​<⋯<jn​≤n∑​u1​u2​⋯uk​∑​v1​v2​⋯vn−k​∑​(−1)τ(i1​i2​⋯ik​i1′​i2′​⋯in−k′​)+τ(u1​u2​⋯uk​v1​v2​⋯vn−k​)ai1​u1​​ai2​u2​​⋯aik​uk​​ai1′​v1​​ai2′​v2​​⋯ain−k′​vn−k​​​

\\quad 要注意到，

a. 对于 $n$ 元排列 i1i2⋯iki1′i2′⋯in−k′i_{1}i_{2}\\cdots i_{k} i_{1}'i_{2}'\\cdots i_{n-k}'i1​i2​⋯ik​i1′​i2′​⋯in−k′​，由于比 i1i_{1}i1​ 小的数有 i1−1i_{1}-1i1​−1 个，比 i2i_{2}i2​ 小的数有 i2−1i_{2}-1i2​−1 个，再除去 i1i_{1}i1​，有 i2−2i_{2}-2i2​−2 个 ……，比 iki_{k}ik​ 小的数有 ik−ki_{k}-kik​−k 个，于是

(−1)i1i2⋯iki1′i2′⋯in−k′=(−1)i1+i2+⋯+ik−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)i1​i2​⋯ik​i1′​i2′​⋯in−k′​=(−1)i1​+i2​+⋯+ik​−2k(k+1)​

b. 假设经过 $S$ 次对换，排列 u1u2⋯ukv1v2⋯vn−ku_{1}u_{2}\\cdots u_{k}v_{1}v_{2}\\cdots v_{n-k}u1​u2​⋯uk​v1​v2​⋯vn−k​ 化为排列 j1j2⋯jkv1v2⋯vn−kj_{1}j_{2}\\cdots j_{k} v_{1}v_{2}\\cdots v_{n-k}j1​j2​⋯jk​v1​v2​⋯vn−k​. 因为 j1<j2<⋯<jkj_{1}<j_{2}<\\cdots<j_{k}j1​<j2​<⋯<jk​，因此 τ(j1⋯jk)=0\\tau(j_{1}\\cdots j_{k})=0τ(j1​⋯jk​)=0，从而

(−1)τ(u1u2⋯uk)=(−1)S(-1)^{\\tau(u_{1}u_{2}\\cdots u_{k})} = (-1)^{S} (−1)τ(u1​u2​⋯uk​)=(−1)S

\\quad 于是

(−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−k(k+1)2⋅(−1)τ(u1u2⋯uk)+τ(v1v2⋯vn−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(u1​u2​⋯uk​v1​v2​⋯vn−k​)​=(−1)S⋅(−1)τ(j1​j2​⋯jk​v1​v2​⋯vn−k​)=(−1)τ(u1​u2​⋯uk​)⋅(−1)τ(j1​j2​⋯jk​v1​v2​⋯vn−k​)=(−1)j1​+j2​+⋯+jk​−2k(k+1)​⋅(−1)τ(u1​u2​⋯uk​)+τ(v1​v2​⋯vn−k​)​

\\quad 从而，

∣A∣=∑1≤j1<j2<⋯<jn≤n∑u1u2⋯uk∑v1v2⋯vn−k(−1)τ(i1i2⋯iki1′i2′⋯in−k′)+τ(u1u2⋯ukv1v2⋯vn−k)ai1u1ai2u2⋯aikukai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jn≤n(−1)τ(i1+i2+⋯+ik)+τ(j1+j2+⋯+jk)∑u1u2⋯uk(−1)τ(u1u2⋯uk)ai1u1ai2u2⋯aikuk∑v1v2⋯vn−k(−1)τ(v1v2⋯vn−k)ai1′v1ai2′v2⋯ain−k′vn−k=∑1≤j1<j2<⋯<jk≤n(−1)(i1+i2+⋯+ik)+(j1+j2+⋯+jk)⋅A(i1,i2,⋯,ikj1,j2,⋯,jk)⋅A(i1′,i2′,⋯,in−k′j1′,j2′,⋯,jn−k′)\\begin{aligned} |A| &= \\sum_{1\\le j_{1} <j_{2}<\\cdots<j_{n}\\le n}\\sum_{u_{1}u_{2}\\cdots u_{k}}\\sum_{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}<j_{2}<\\cdots<j_{n}\\le n}(-1)^{\\tau(i_{1}+i_{2} +\\cdots + i_{k}) + \\tau(j_{1} + j_{2} + \\cdots + j_{k})} \\sum_{u_{1}u_{2}\\cdots u_{k}} (-1)^{\\tau(u_{1}u_{2}\\cdots u_{k})} a_{i_{1}u_{1}}a_{i_{2}u_{2}}\\cdots a_{i_{k}u_{k}} \\sum_{v_{1}v_{2}\\cdots v_{n-k}}(-1)^{\\tau(v_{1}v_{2}\\cdots v_{n-k})}a_{i_{1}'v_{1}}a_{i_{2}'v_{2}}\\cdots a_{i_{n-k}'v_{n-k}}\\\\ &= \\sum_{1\\le j_{1}<j_{2}<\\cdots<j_{k}\\le n} (-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_{k}\\\\ j_{1},j_{2},\\cdots,j_{k}\\end{matrix}\\right) \\cdot A\\left(\\begin{matrix} i_{1}',i_{2}',\\cdots,i_{n-k}'\\\\ j_{1}',j_{2}',\\cdots,j_{n-k}'\\end{matrix}\\right) \\end{aligned} ∣A∣​=1≤j1​<j2​<⋯<jn​≤n∑​u1​u2​⋯uk​∑​v1​v2​⋯vn−k​∑​(−1)τ(i1​i2​⋯ik​i1′​i2′​⋯in−k′​)+τ(u1​u2​⋯uk​v1​v2​⋯vn−k​)ai1​u1​​ai2​u2​​⋯aik​uk​​ai1′​v1​​ai2′​v2​​⋯ain−k′​vn−k​​=1≤j1​<j2​<⋯<jn​≤n∑​(−1)τ(i1​+i2​+⋯+ik​)+τ(j1​+j2​+⋯+jk​)u1​u2​⋯uk​∑​(−1)τ(u1​u2​⋯uk​)ai1​u1​​ai2​u2​​⋯aik​uk​​v1​v2​⋯vn−k​∑​(−1)τ(v1​v2​⋯vn−k​)ai1′​v1​​ai2′​v2​​⋯ain−k′​vn−k​​=1≤j1​<j2​<⋯<jk​≤n∑​(−1)(i1​+i2​+⋯+ik​)+(j1​+j2​+⋯+jk​)⋅A(i1​,i2​,⋯,ik​j1​,j2​,⋯,jk​​)⋅A(i1′​,i2′​,⋯,in−k′​j1′​,j2′​,⋯,jn−k′​​)​

#

\\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∣=∣a11⋯a1k0⋯0⋮⋮⋮⋮ak1⋯akk0⋯0c11⋯c1kb11⋯b1t⋮⋮⋮⋮ct1⋯ctkbt1⋯btt∣|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​⋮ak1​c11​⋮ct1​​⋯⋯⋯⋯​a1k​⋮akk​c1k​⋮ctk​​0⋮0b11​⋮bt1​​⋯⋯⋯⋯​0⋮0b1t​⋮btt​​​