《大学数学3（第三版）》第一章（2）

文章目录

• • • 第四节|行列式按行（列）展开
    • • 余子式与代数余子式
      • 拉普拉斯展开定理
      • [例题 1 1 1](#例题 1 1 1)
      • 拉普拉斯定理的推广
      • [例题 2 2 2](#例题 2 2 2)
    • 第五节|克拉默法则
    • • 克拉默法则

第四节|行列式按行（列）展开

余子式与代数余子式

• 余子式：在 n n n阶行列式中，把元素 a i j a_{ij} aij（ i i i， j = 1 j=1 j=1， 2 2 2， ⋯ \cdots ⋯， n n n）所在的行和列划去后，剩下的 ( n − 1 ) 2 (n - 1)^{2} (n−1)2个元素按原来的顺序构成的 n − 1 n - 1 n−1阶行列式称为元素 a i j a_{ij} aij的余子式，记作 M i j M_{ij} Mij
• 代数余子式：余子式带上符号 ( − 1 ) i + j (-1)^{i + j} (−1)i+j称为 a i j a_{ij} aij的代数余子式，记作 A i j = ( − 1 ) i + j M i j A_{ij}=(-1)^{i + j}M_{ij} Aij=(−1)i+jMij

拉普拉斯展开定理

• n n n阶行列式等于它的任一行（列）的各元素与其对应的代数余子式乘积之和，即

D = ∑ k = 1 n a i k A i k , i = 1 , 2 , ⋯   , n D= \sum_{k = 1}^{n}a_{ik}A_{ik} , \ \ \ \ i = 1,2,\cdots,n D=k=1∑naikAik, i=1,2,⋯,n

• • 证明
    • 先证

D ′ = ∣ a 11 0 ⋯ 0 a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = a 11 A 11 D^{'}= \begin{vmatrix} a_{11} & 0 & \cdots & 0 \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}= a_{11}A_{11} D′= a11a21⋮an10a22⋮an2⋯⋯⋯0a2n⋮ann =a11A11

• • • 由行列式定义

∑ ( j 1 j 2 ⋯ j n ) ( − 1 ) τ ( j 1 j 2 ⋯ j n ) a 1 j 1 a 2 j 2 ⋯ a n j n = ∑ ( j 2 ⋯ j n ) ( − 1 ) τ ( 1 j 2 ⋯ j n ) a 11 a 2 j 2 ⋯ a n j n = a 11 ∑ ( j 2 ⋯ j n ) ( − 1 ) τ ( j 2 ⋯ j n ) a 2 j 2 ⋯ a n j n = a 11 M 11 = a 11 A 11 \begin{aligned} \sum_{(j_{1}j_{2} \cdots j_{n})}(-1)^{\tau(j_{1}j_{2} \cdots j_{n})}a_{1j_{1}}a_{2j_{2}} \cdots a_{nj_{n}} &= \sum_{(j_{2} \cdots j_{n})}(-1)^{\tau(1j_{2} \cdots j_{n})}a_{11}a_{2j_{2}} \cdots a_{nj_{n}} \\ &= a_{11}\sum_{(j_{2} \cdots j_{n})}(-1)^{\tau(j_{2} \cdots j_{n})}a_{2j_{2}} \cdots a_{nj_{n}} \\ &= a_{11}M_{11} \\ &= a_{11}A_{11} \end{aligned} (j1j2⋯jn)∑(−1)τ(j1j2⋯jn)a1j1a2j2⋯anjn=(j2⋯jn)∑(−1)τ(1j2⋯jn)a11a2j2⋯anjn=a11(j2⋯jn)∑(−1)τ(j2⋯jn)a2j2⋯anjn=a11M11=a11A11

• • • 再证一般情形

D ′ ′ = ∣ a 11 ⋯ a 1 j ⋯ a 1 n ⋮ ⋮ ⋮ 0 ⋯ a i j ⋯ 0 ⋮ ⋮ ⋮ a n 1 ⋯ a n j ⋯ a n n ∣ = a i j A i j D^{''}= \begin{vmatrix} a_{11} & \cdots & a_{1j} & \cdots & a_{1n} \\ \vdots & & \vdots & & \vdots \\ 0 & \cdots & a_{ij} & \cdots & 0 \\ \vdots & & \vdots & & \vdots \\ a_{n1} & \cdots & a_{nj} & \cdots & a_{nn} \\ \end{vmatrix}= a_{ij}A_{ij} D′′= a11⋮0⋮an1⋯⋯⋯a1j⋮aij⋮anj⋯⋯⋯a1n⋮0⋮ann =aijAij

• • • 把 D ′ ′ D^{''} D′′的第 i i i行依次与第 i − 1 i - 1 i−1， ⋯ \cdots ⋯， 2 2 2， 1 1 1行交换后换到第一行，再把第 j j j列依次与 j − 1 j - 1 j−1， ⋯ \cdots ⋯， 2 2 2， 1 1 1列交换后换到第 1 1 1列，则总共经过 i + j − 2 i + j - 2 i+j−2次交换后，把 a i j a_{ij} aij交换到 D ′ ′ D^{''} D′′的左上角，化成 D ′ D^{'} D′的形式，从而

D ′ ′ = ( − 1 ) i + j − 2 a i j M i j = a i j ( − 1 ) i + j M i j = a i j A i j D^{''}= (-1)^{i + j - 2}a_{ij}M_{ij}= a_{ij}(-1)^{i + j}M_{ij}= a_{ij}A_{ij} D′′=(−1)i+j−2aijMij=aij(−1)i+jMij=aijAij

D = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 + 0 + ⋯ + 0 0 + a i 2 + 0 + ⋯ + 0 ⋯ 0 + ⋯ + 0 + a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 0 ⋯ 0 ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ + ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ 0 a i 2 ⋯ 0 ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ + ⋯ + ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ 0 0 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ = a i 1 A i 1 + a i 2 A i 2 + ⋯ + a i n A i n ( i = 1 , 2 , ⋯   , n ) \begin{aligned} D &= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} + 0 + \cdots + 0 & 0 + a_{i2} + 0 + \cdots + 0 & \cdots & 0 + \cdots + 0 + a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ &= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ 0 & a_{i2} & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \cdots + \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \\ &= a_{i1}A_{i1} + a_{i2}A_{i2} + \cdots + a_{in}A_{in}(i = 1,2,\cdots,n) \end{aligned} D= a11⋮ai1+0+⋯+0⋮an1a12⋮0+ai2+0+⋯+0⋮an2⋯⋯⋯a1n⋮0+⋯+0+ain⋮ann = a11⋮ai1⋮an1a12⋮0⋮an2⋯⋯⋯a1n⋮0⋮ann + a11⋮0⋮an1a12⋮ai2⋮an2⋯⋯⋯a1n⋮0⋮ann +⋯+ a11⋮0⋮an1a12⋮0⋮an2⋯⋯⋯a1n⋮ain⋮ann =ai1Ai1+ai2Ai2+⋯+ainAin(i=1,2,⋯,n)

• n n n阶行列式的任一行（列）的各元素与另一行（列）对应元素的代数余子式的乘积之和等于零，即

∑ k = 1 n a i k A j k = 0 , i ≠ j \sum_{k = 1}^{n}a_{ik}A_{jk} = 0 , \ \ \ \ i \neq j k=1∑naikAjk=0, i=j

• • 证明

D = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a j 1 a j 2 ⋯ a j n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{j1} & a_{j2} & \cdots & a_{jn} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D= a11⋮ai1⋮aj1⋮an1a12⋮ai2⋮aj2⋮an2⋯⋯⋯⋯a1n⋮ain⋮ajn⋮ann

D 1 = ∣ a 11 a 12 ⋯ a 1 n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a i 1 a i 2 ⋯ a i n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ D_{1}= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{i1} & a_{i2} & \cdots & a_{in} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} D1= a11⋮ai1⋮ai1⋮an1a12⋮ai2⋮ai2⋮an2⋯⋯⋯⋯a1n⋮ain⋮ain⋮ann

• • • 显然， D 1 = 0 D_{1} = 0 D1=0，且 D 1 D_{1} D1与 D D D的第 j j j行各元素的代数余子式对应相等，将 D 1 D_{1} D1按第 j j j行展开，得

a i 1 A j 1 + a i 2 A j 2 + ⋯ + a i n A j n = ∑ k = 1 n a i k A j k = 0 a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}= \sum_{k = 1}^{n}a_{ik}A_{jk} = 0 ai1Aj1+ai2Aj2+⋯+ainAjn=k=1∑naikAjk=0

• • • 即 D D D中第 i i i行各元素与第 j j j行对应元素的代数余子式得乘积之和等于零

例题 1 1 1

• 问题：证明范德蒙德行列式

D n = ∣ 1 1 ⋯ 1 x 1 x 2 ⋯ x n x 1 2 x 2 2 ⋯ x n 2 ⋮ ⋮ ⋮ x 1 n − 1 x 2 n − 1 ⋯ x n n − 1 ∣ = ∏ 1 ≤ j < i ≤ n ( x i − x j ) , n ≥ 2 D_{n}= \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_{1} & x_{2} & \cdots & x_{n} \\ x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2} \\ \vdots & \vdots & & \vdots \\ x_{1}^{n - 1} & x_{2}^{n - 1} & \cdots & x_{n}^{n - 1} \end{vmatrix}= \prod_{1 \leq j < i \leq n}(x_{i} - x_{j}) , \ \ \ \ n \geq 2 Dn= 1x1x12⋮x1n−11x2x22⋮x2n−1⋯⋯⋯⋯1xnxn2⋮xnn−1 =1≤j<i≤n∏(xi−xj), n≥2

• 解答
  • 用数学归纳法证明
  • 当 n = 2 n = 2 n=2时

D 2 = ∣ 1 1 x 1 x 2 ∣ = x 2 − x 1 D_{2}= \begin{vmatrix} 1 & 1 \\ x_{1} & x_{2} \end{vmatrix}= x_{2} - x_{1} D2= 1x11x2 =x2−x1

• • 结论成立
  • 假设对于 n − 1 n - 1 n−1阶范德蒙德行列式结论成立，证 n n n阶情形，对于 D n D_{n} Dn，从第 n n n行开始，每行减去上面一行与 x 1 x_{1} x1的乘积，得

D n = ∣ 1 1 ⋯ 1 0 x 2 − x 1 ⋯ x n − x 1 0 x 2 ( x 2 − x 1 ) ⋯ x n ( x n − x 1 ) ⋮ ⋮ ⋮ 0 x 2 n − 2 ( x 2 − x 1 ) ⋯ x n n − 2 ( x n − x 1 ) ∣ = ∣ x 2 − x 1 ⋯ x n − x 1 x 2 ( x 2 − x 1 ) ⋯ x n ( x n − x 1 ) ⋮ ⋮ x 2 n − 2 ( x 2 − x 1 ) ⋯ x n n − 2 ( x n − x 1 ) ∣ = ( x 2 − x 1 ) ⋯ ( x n − x 1 ) ∣ 1 1 ⋯ 1 x 2 x 3 ⋯ x n x 2 2 x 3 2 ⋯ x n 2 ⋮ ⋮ ⋮ x 2 n − 2 x 3 n − 2 ⋯ x n n − 2 ∣ \begin{aligned} D_{n} &= \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & x_{2} - x_{1} & \cdots & x_{n} - x_{1} \\ 0 & x_{2}(x_{2} - x_{1}) & \cdots & x_{n}(x_{n} - x_{1}) \\ \vdots & \vdots & & \vdots \\ 0 & x_{2}^{n - 2}(x_{2} - x_{1}) & \cdots & x_{n}^{n - 2}(x_{n} - x_{1}) \end{vmatrix} \\ &= \begin{vmatrix} x_{2} - x_{1} & \cdots & x_{n} - x_{1} \\ x_{2}(x_{2} - x_{1}) & \cdots & x_{n}(x_{n} - x_{1}) \\ \vdots & & \vdots \\ x_{2}^{n - 2}(x_{2} - x_{1}) & \cdots & x_{n}^{n - 2}(x_{n} - x_{1}) \end{vmatrix} \\ &= (x_{2} - x_{1}) \cdots (x_{n} - x_{1}) \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_{2} & x_{3} & \cdots & x_{n} \\ x_{2}^{2} & x_{3}^{2} & \cdots & x_{n}^{2} \\ \vdots & \vdots & & \vdots \\ x_{2}^{n - 2} & x_{3}^{n - 2} & \cdots & x_{n}^{n - 2} \end{vmatrix} \end{aligned} Dn= 100⋮01x2−x1x2(x2−x1)⋮x2n−2(x2−x1)⋯⋯⋯⋯1xn−x1xn(xn−x1)⋮xnn−2(xn−x1) = x2−x1x2(x2−x1)⋮x2n−2(x2−x1)⋯⋯⋯xn−x1xn(xn−x1)⋮xnn−2(xn−x1) =(x2−x1)⋯(xn−x1) 1x2x22⋮x2n−21x3x32⋮x3n−2⋯⋯⋯⋯1xnxn2⋮xnn−2

• • 上式右端得行列式为 n − 1 n - 1 n−1阶范德蒙德行列式，于是由归纳假设有

D n = ( x 2 − x 1 ) ⋯ ( x n − x 1 ) ∏ 2 ≤ j < i ≤ n ( x i − x j ) = ∏ 1 ≤ j < i ≤ n ( x i − x j ) \begin{aligned} D_{n} &= (x_{2} - x_{1}) \cdots (x_{n} - x_{1})\prod_{2 \leq j < i \leq n}(x_{i} - x_{j}) \\ &= \prod_{1 \leq j < i \leq n}(x_{i} - x_{j}) \end{aligned} Dn=(x2−x1)⋯(xn−x1)2≤j<i≤n∏(xi−xj)=1≤j<i≤n∏(xi−xj)

拉普拉斯定理的推广

• 在 n n n阶行列式 D D D中，任意选定 k k k行 k k k列（ 1 ≤ k ≤ n 1 \leq k \leq n 1≤k≤n），对于这些行和列交叉处的 k 2 k^{2} k2个元素，按原来的顺序构成一个 k k k阶行列式 M M M，称为 D D D的一个 k k k阶子式，划去这 k k k行 k k k列，余下的元素按原来的顺序构成一个 n − k n - k n−k阶行列式，在其前面冠以符号 ( − 1 ) i 1 + i 2 + ⋯ + i k + j 1 + j 2 + ⋯ + j k (-1)^{i_{1} + i_{2} + \cdots + i_{k} + j_{1} + j_{2} + \cdots + j_{k}} (−1)i1+i2+⋯+ik+j1+j2+⋯+jk称为 M M M的代数余子式
• 在 n n n阶行列式 D D D中，任意取定 k k k行（列）（ 1 ≤ k ≤ n − 1 1 \leq k \leq n - 1 1≤k≤n−1），由这 k k k行（列）组成的所有 k k k阶子式与它们的代数余子式的乘积之和等于行列式 D D D

例题 2 2 2

• 问题：证明

D = ∣ a 11 ⋯ a 1 m 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ a m 1 ⋯ a m m 0 ⋯ 0 c 11 ⋯ c 1 m b 11 ⋯ b 1 n ⋮ ⋮ ⋮ ⋮ c n 1 ⋯ c n m b n 1 ⋯ b n n ∣ = ∣ a 11 ⋯ a 1 m ⋮ ⋮ a m 1 ⋯ a m m ∣ ∣ b 11 ⋯ b 1 n ⋮ ⋮ b n 1 ⋯ b n n ∣ D= \begin{vmatrix} a_{11} & \cdots & a_{1m} & 0 & \cdots & 0 \\ \vdots & & \vdots & \vdots & & \vdots \\ a_{m1} & \cdots & a_{mm} & 0 & \cdots & 0 \\ c_{11} & \cdots & c_{1m} & b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots & \vdots & & \vdots \\ c_{n1} & \cdots & c_{nm} & b_{n1} & \cdots & b_{nn} \\ \end{vmatrix}= \begin{vmatrix} a_{11} & \cdots & a_{1m}\\ \vdots & & \vdots\\ a_{m1} & \cdots & a_{mm} \end{vmatrix} \begin{vmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{vmatrix} D= a11⋮am1c11⋮cn1⋯⋯⋯⋯a1m⋮ammc1m⋮cnm0⋮0b11⋮bn1⋯⋯⋯⋯0⋮0b1n⋮bnn = a11⋮am1⋯⋯a1m⋮amm b11⋮bn1⋯⋯b1n⋮bnn

• 解答：取前面的 m m m行，由这 m m m行组成的所有 m m m阶子式中只有 D 1 = ∣ a 11 ⋯ a 1 m ⋮ ⋮ a m 1 ⋯ a m m ∣ D_{1} = \begin{vmatrix} a_{11} & \cdots & a_{1m}\\ \vdots & & \vdots\\ a_{m1} & \cdots & a_{mm} \end{vmatrix} D1= a11⋮am1⋯⋯a1m⋮amm 可能不为 0 0 0，其他的子式全为 0 0 0，所以行列式的值等于 D 1 D_{1} D1乘它的代数余子式，即

D = D 1 ( − 1 ) ( 1 + 2 + ⋯ + m ) + ( 1 + 2 + ⋯ + m ) ∣ b 11 ⋯ b 1 n ⋮ ⋮ b n 1 ⋯ b n n ∣ = ∣ a 11 ⋯ a 1 m ⋮ ⋮ a m 1 ⋯ a m m ∣ ∣ b 11 ⋯ b 1 n ⋮ ⋮ b n 1 ⋯ b n n ∣ D= D_{1}(-1)^{(1 + 2 + \cdots + m) + (1 + 2 + \cdots + m)} \begin{vmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{vmatrix}= \begin{vmatrix} a_{11} & \cdots & a_{1m}\\ \vdots & & \vdots\\ a_{m1} & \cdots & a_{mm} \end{vmatrix} \begin{vmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & & \vdots \\ b_{n1} & \cdots & b_{nn} \end{vmatrix} D=D1(−1)(1+2+⋯+m)+(1+2+⋯+m) b11⋮bn1⋯⋯b1n⋮bnn = a11⋮am1⋯⋯a1m⋮amm b11⋮bn1⋯⋯b1n⋮bnn

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第五节|克拉默法则

克拉默法则

• 如果含有 n n n个方程的 n n n元线性方程组

{ a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2 ⋯ a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n = b n (1) \begin{cases} a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n} = b_{1} \\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} = b_{2} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \\ a_{n1}x_{1} + a_{n2}x_{2} + \cdots + a_{nn}x_{n} = b_{n} \end{cases} \tag{1} ⎩ ⎨ ⎧a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2 ⋯an1x1+an2x2+⋯+annxn=bn(1)

• 其系数行列式

D = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ ≠ 0 D= \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} \neq 0 D= a11a21⋮an1a12a22⋮an2⋯⋯⋯a1na2n⋮ann =0

• 则线性方程组（ 1 1 1）有唯一解，且其解可表示为

x 1 = D 1 D , x 2 = D 2 D , ⋯   , x n = D n D (2) x_{1} = \frac{D_{1}}{D},x_{2} = \frac{D_{2}}{D},\cdots,x_{n} = \frac{D_{n}}{D} \tag{2} x1=DD1,x2=DD2,⋯,xn=DDn(2)

D j = ∣ a 11 ⋯ a 1 , j − 1 b 1 a 1 , j + 1 ⋯ a 1 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 ⋯ a n , j − 1 b n a n , j + 1 ⋯ a n n ∣ , j = 1 , 2 , ⋯   , n D_{j}= \begin{vmatrix} a_{11} & \cdots & a_{1,j - 1} & b_{1} & a_{1,j + 1} & \cdots & a_{1n} \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ a_{n1} & \cdots & a_{n,j - 1} & b_{n} & a_{n,j + 1} & \cdots & a_{nn} \end{vmatrix} , \ \ \ \ j = 1,2,\cdots,n Dj= a11⋮an1⋯⋯a1,j−1⋮an,j−1b1⋮bna1,j+1⋮an,j+1⋯⋯a1n⋮ann , j=1,2,⋯,n

• 证明

D j = ∑ k = 1 n b k A k j D_{j}= \sum_{k = 1}^{n}b_{k}A_{kj} Dj=k=1∑nbkAkj

• • 首先证明由（ 2 2 2）式确定的 x 1 , x 2 , ⋯   , x n x_{1},x_{2},\cdots,x_{n} x1,x2,⋯,xn为线性方程组（1）的解，将 x 1 , x 2 , ⋯   , x n x_{1},x_{2},\cdots,x_{n} x1,x2,⋯,xn代入线性方程组（1）的第 i i i个方程（ i = 1 , 2 , ⋯   , n i = 1,2,\cdots,n i=1,2,⋯,n得

a i 1 x 1 + a i 2 x 2 + ⋯ + a i n x n = ∑ j = 1 n a i j x j = ∑ j = 1 n a i j D j D = 1 D ∑ j = 1 n a i j ( ∑ k = 1 n b k A k j ) = 1 D ∑ k = 1 n b k ( ∑ j = 1 n a i j A k j ) = b i D ∑ j = 1 n a i j A i j = b i D ⋅ D = b i \begin{aligned} a_{i1}x_{1} + a_{i2}x_{2} + \cdots + a_{in}x_{n} &= \sum_{j = 1}^{n}a_{ij}x_{j}= \sum_{j = 1}^{n}a_{ij}\frac{D_{j}}{D} \\ &= \frac{1}{D}\sum_{j = 1}^{n}a_{ij}(\sum_{k = 1}^{n}b_{k}A_{kj}) \\ &= \frac{1}{D}\sum_{k = 1}^{n}b_{k}(\sum_{j = 1}^{n}a_{ij}A_{kj}) \\ &= \frac{b_{i}}{D}\sum_{j = 1}{n}a_{ij}A_{ij} \\ &= \frac{b_{i}}{D} \cdot D \\ &= b_{i} \end{aligned} ai1x1+ai2x2+⋯+ainxn=j=1∑naijxj=j=1∑naijDDj=D1j=1∑naij(k=1∑nbkAkj)=D1k=1∑nbk(j=1∑naijAkj)=Dbij=1∑naijAij=Dbi⋅D=bi

• • 即 x 1 , x 2 , ⋯   , x n x_{1},x_{2},\cdots,x_{n} x1,x2,⋯,xn为线性方程组（ 1 1 1）的解
  • 然后证明解的唯一性，若 x 1 , x 2 , ⋯   , x n x_{1},x_{2},\cdots,x_{n} x1,x2,⋯,xn为线性方程组（ 1 1 1）的解，用行列式 D D D的第 j j j列各元素的代数余子式 A 1 j , A 2 j , ⋯   , A n j A_{1j},A_{2j},\cdots,A_{nj} A1j,A2j,⋯,Anj分别乘以线性方程组（ 1 1 1）的第 1 1 1个，第 2 2 2个， ⋯ \cdots ⋯，第 n n n个方程并相加得

( ∑ k = 1 n a k 1 A k j ) x 1 + ⋯ + ( ∑ k = 1 n a k j A k j ) x j + ⋯ + ( ∑ k = 1 n a k n A k j ) x n = ∑ k = 1 n b k A k j (\sum_{k = 1}{n}a_{k1}A_{kj})x_{1} + \cdots + (\sum_{k = 1}^{n}a_{kj}A_{kj})x_{j} + \cdots + (\sum_{k = 1}^{n}a_{kn}A_{kj})x_{n} = \sum_{k = 1}^{n}b_{k}A_{kj} (k=1∑nak1Akj)x1+⋯+(k=1∑nakjAkj)xj+⋯+(k=1∑naknAkj)xn=k=1∑nbkAkj

D x j = D j , j = 1 , 2 , ⋯   , n Dx_{j} = D_{j} , \ \ \ \ j = 1,2,\cdots,n Dxj=Dj, j=1,2,⋯,n

• • 由于 D ≠ 0 D \neq 0 D=0， x j = D j D ( j = 1 , 2 , ⋯   , n ) x_{j} = \frac{D_{j}}{D}(j = 1,2,\cdots,n) xj=DDj(j=1,2,⋯,n)，从而线性方程组（ 1 1 1）的解 x 1 , x 2 , ⋯   , x n x_{1},x_{2},\cdots,x_{n} x1,x2,⋯,xn满足（ 2 2 2）

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