行列式
利用预先定义的计算规则，根据方阵计算出来的值。有点像列竖式计算的竖式，本质上是一个数值。
二阶行列式计算：
∣ a b c d ∣ = a d − b c \begin{vmatrix}
a & b \\
c & d
\end{vmatrix} = ad - bc a c b d = a d − b c 三阶行列式的计算：
∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ = a 11 a 22 a 33 + a 12 a 23 a 31 + a 21 a 32 a 13 − a 12 a 21 a 33 − a 23 a 32 a 11 − a 13 a 22 a 31 \begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{vmatrix} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{21}a_{32}a_{13} - a_{12}a_{21}a_{33} - a_{23}a_{32}a_{11} - a_{13}a_{22}a_{31} a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 = a 11 a 22 a 33 + a 12 a 23 a 31 + a 21 a 32 a 13 − a 12 a 21 a 33 − a 23 a 32 a 11 − a 13 a 22 a 31 预先定义的计算规则为：所有不同行不同列的元素的乘积的代数和。每一项乘积的符号由逆序数 τ ( j 1 j 2 … j n ) \tau(j_1 j_2 \dots j_n) τ ( j 1 j 2 … j n )
n n n
D = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∑ j 1 j 2 … j n ( − 1 ) τ ( j 1 j 2 … j n ) a 1 j 1 a 2 j 2 a 3 j 3 ⋯ a n j n D =
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix} = \sum_{j_1 j_2 \dots j_n} (-1)^{\tau(j_1 j_2 \dots j_n)} a_{1j_1} a_{2j_2} a_{3j_3} \cdots a_{nj_n} D = a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋱ ⋯ a 1 n a 2 n ⋮ a nn = j 1 j 2 … j n ∑ ( − 1 ) τ ( j 1 j 2 … j n ) a 1 j 1 a 2 j 2 a 3 j 3 ⋯ a n j n 行列式的性质
D T = D D^{T} = D D T = D 如果 D D D 0 0 0 D = 0 D = 0 D = 0
行列变换法则：
性质 1 交换两行（列），符号改变
∣ a 11 a 12 ⋯ a 1 i ⋯ a 1 j ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ a 2 j ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ a n j ⋯ a n n ∣ = − ∣ a 11 a 12 ⋯ a 1 j ⋯ a 1 i ⋯ a 1 n a 21 a 22 ⋯ a 2 j ⋯ a 2 i ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n j ⋯ a n i ⋯ a n n ∣ \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & a_{1j} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & a_{2j} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & a_{nj} & \cdots & a_{nn} \\
\end{vmatrix} = -
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1j} & \cdots & a_{1i} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2j} & \cdots & a_{2i} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nj} & \cdots & a_{ni} & \cdots & a_{nn} \\
\end{vmatrix} a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 j a 2 j ⋮ a nj ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn = − a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 j a 2 j ⋮ a nj ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn 推论 1 行列式两行（列）相等，值为 0
∣ a 11 a 12 ⋯ a 1 i ⋯ a 1 i ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ a 2 i ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ a n i ⋯ a n n ∣ = 0 \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & a_{1i} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & a_{2i} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & a_{ni} & \cdots & a_{nn} \\
\end{vmatrix} = 0 a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn = 0
证明 ：
交换相等的两行，符号改变，但是行列式还是原来的行列式，所以为 0。
性质 2 行列式等于将某一行拆成两个行列式相加
∣ a 11 a 12 ⋯ a 1 i ⋯ b 1 i + a 1 j ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ b 2 i + a 2 j ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ b n i + a n j ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ a 1 i ⋯ b 1 i ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ b 2 i ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ b n i ⋯ a n n ∣ + ∣ a 11 a 12 ⋯ a 1 i ⋯ a 1 j ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ a 2 j ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ a n j ⋯ a n n ∣ \begin{aligned}
& \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & b_{1i} + a_{1j} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & b_{2i} + a_{2j} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & b_{ni} + a_{nj} & \cdots & a_{nn} \\
\end{vmatrix} \\
& = \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & b_{1i} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & b_{2i} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & b_{ni} & \cdots & a_{nn} \\
\end{vmatrix} +
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & a_{1j} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & a_{2j} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & a_{nj} & \cdots & a_{nn} \\
\end{vmatrix}
\end{aligned} a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ b 1 i + a 1 j b 2 i + a 2 j ⋮ b ni + a nj ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn = a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ b 1 i b 2 i ⋮ b ni ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn + a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 j a 2 j ⋮ a nj ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn 推论 1 两行（列）相加，值不变
∣ a 11 a 12 ⋯ a 1 i ⋯ a 1 j ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ a 2 j ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ a n j ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ a 1 i ⋯ a 1 j + a 1 i ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ a 2 j + a 2 i ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ a n j + a n i ⋯ a n n ∣ \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & a_{1j} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & a_{2j} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & a_{nj} & \cdots & a_{nn} \\
\end{vmatrix} =
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & a_{1j} + a_{1i} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & a_{2j} + a_{2i} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & a_{nj} + a_{ni} & \cdots & a_{nn} \\
\end{vmatrix} a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 j a 2 j ⋮ a nj ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn = a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ a 1 j + a 1 i a 2 j + a 2 i ⋮ a nj + a ni ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn
证明 ：
将后一个行列式拆成两个行列式的和，然后因为后一个行列式的两列式相等的，由性质 1 得到等于 0.
性质 3 行列式乘一个数，等于行列式某一行（列）乘一个数
k ∣ a 11 a 12 ⋯ a 1 i ⋯ ⋯ a 1 n a 21 a 22 ⋯ a 2 i ⋯ ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n i ⋯ ⋯ a n n ∣ = ∣ a 11 a 12 ⋯ k a 1 i ⋯ a 1 n a 21 a 22 ⋯ k a 2 i ⋯ a 2 n ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ k a n i ⋯ a n n ∣ k
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1i} & \cdots & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2i} & \cdots & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{ni} & \cdots & \cdots & a_{nn} \\
\end{vmatrix} =
\begin{vmatrix}
a_{11} & a_{12} & \cdots & ka_{1i} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & ka_{2i} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & ka_{ni} & \cdots & a_{nn} \\
\end{vmatrix} k a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 i a 2 i ⋮ a ni ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn = a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ k a 1 i k a 2 i ⋮ k a ni ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn 活用以上三条性质及其推论可以得到更好用的性质。
我们引入以下记号表达相应的含义：
r i ÷ k r_{i} \div k r i ÷ k c i ÷ k c_{i} \div k c i ÷ k i i i k k k r i + k r j r_{i} + kr_{j} r i + k r j c i + k r j c_{i} + kr_{j} c i + k r j j j j k k k i i i r i ↔ r j r_{i} \leftrightarrow r_{j} r i ↔ r j c i ↔ c j c_{i} \leftrightarrow c_{j} c i ↔ c j i i i j j j
按行（列）展开
为了方便表示，我们定义一些概念：
余子式 M i j M_{ij} M ij i i i j j j
M i j = ∣ a 11 ⋯ a 1 , j − 1 a 1 , j + 1 ⋯ a 1 n a 21 ⋯ a 2 , j − 1 a 1 , j + 1 ⋯ a 2 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 ∣ M_{ij} =
\begin{vmatrix}
a_{11} & \cdots & a_{1, j-1} & a_{1, j+1} & \cdots & a_{1n} \\
a_{21} & \cdots & a_{2, j-1} & a_{1, j+1} & \cdots & a_{2n} \\
\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_{nj-1} & a_{n, j+1} & \cdots & a_{nn} \\
\end{vmatrix} M ij = a 11 a 21 ⋮ a i − 1 , 1 a i − 1 , 1 ⋮ a n 1 ⋯ ⋯ ⋯ ⋯ ⋯ a 1 , j − 1 a 2 , j − 1 ⋮ a i − 1 , j − 1 a i − 1 , j − 1 ⋮ a nj − 1 a 1 , j + 1 a 1 , j + 1 ⋮ a i − 1 , j + 1 a i − 1 , j + 1 ⋮ a n , j + 1 ⋯ ⋯ ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a i 1 , n a i 1 , n ⋮ a nn
代数余子式 A i j A_{ij} A ij ( − 1 ) i + j (-1)^{i+j} ( − 1 ) i + j
于是我们有按行展开 ：D = ∑ j = 1 n a i j A i j D = \sum_{j = 1}^{n} a_{ij}A_{ij} D = ∑ j = 1 n a ij A ij i i i D D D i i i
关于符号项的确定 > 首先对于有 a 11 a_{11} a 11 ( − 1 ) τ ( j 1 j 2 … j n ) = ( − 1 ) τ ( 1 j 2 … j n ) (-1)^{\tau(j_1 j_2 \dots j_n)} = (-1)^{\tau(1 j_2 \dots j_n)} ( − 1 ) τ ( j 1 j 2 … j n ) = ( − 1 ) τ ( 1 j 2 … j n ) a i j a_{ij} a ij a i j a_{ij} a ij a 11 a_{11} a 11 i − 1 i-1 i − 1 j − 1 j-1 j − 1 ( − 1 ) ( i − 1 ) + ( j − 1 ) = ( − 1 ) i + j (-1)^{(i-1) + (j-1)} = (-1)^{i + j} ( − 1 ) ( i − 1 ) + ( j − 1 ) = ( − 1 ) i + j a i j a_{ij} a ij ( − 1 ) i + j × ( − 1 ) τ ( 1 j 2 … j n ) (-1)^{i+j} \times (-1)^{\tau(1 j_2 \dots j_n)} ( − 1 ) i + j × ( − 1 ) τ ( 1 j 2 … j n ) 暂时没有想到合适的说法
克拉默法则
将行列式和方程组联系起来的法则。
已知一个由 n n n n n n
{ a 11 x 11 + a 12 x 12 + ⋯ + a 1 n x 1 n = b 1 , a 21 x 21 + a 22 x 22 + ⋯ + a 2 n x 2 n = b 2 , … a n 1 x n 1 + a n 2 x n 2 + ⋯ + a n n x n n = b n . \begin{cases}
a_{11}x_{11} + a_{12}x_{12} + \cdots + a_{1n}x_{1n} = b_1, \\
a_{21}x_{21} + a_{22}x_{22} + \cdots + a_{2n}x_{2n} = b_2, \\
\dots \\
a_{n1}x_{n1} + a_{n2}x_{n2} + \cdots + a_{nn}x_{nn} = b_n. \\
\end{cases} ⎩ ⎨ ⎧ a 11 x 11 + a 12 x 12 + ⋯ + a 1 n x 1 n = b 1 , a 21 x 21 + a 22 x 22 + ⋯ + a 2 n x 2 n = b 2 , … a n 1 x n 1 + a n 2 x n 2 + ⋯ + a nn x nn = b n . 令
D = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a n 1 a n 2 ⋯ a n n ∣ 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} D = a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a nn 为系数行列式，若 D ≠ 0 D \neq 0 D = 0
x 1 = D 1 D , x 2 = D 2 D , ⋯ , x n = D n D x_1 = \frac{D_1}{D}, x_2 = \frac{D_2}{D}, \cdots , x_n = \frac{D_n}{D} x 1 = D D 1 , x 2 = D D 2 , ⋯ , x n = D D n 其中 D i D_i D i b 1 , b 2 , b 3 , ⋯ , b n b_1, b_2, b_3, \cdots, b_n b 1 , b 2 , b 3 , ⋯ , b n D D D i i i
拉普拉斯展开
可以理解为按块展开，在后来的矩阵分块 中有重要应用。
先定义一些概念：
选定 k k k k k k k 2 k^2 k 2 M M M D D D k k k M M M 余子式 N N N 。 A i j = ( − 1 ) i 1 + i 2 + ⋯ + i k + j 1 + j 2 + ⋯ + j n N i j A_{ij} = (-1)^{i_1 + i_2 + \dots + i_k + j_1 + j_2 + \dots + j_n}N_{ij} A ij = ( − 1 ) i 1 + i 2 + ⋯ + i k + j 1 + j 2 + ⋯ + j n N ij M M M 代数余子式 。
再来介绍拉普拉斯展开：
选定行列式 D D D k k k t = C n k t = C_{n}^{k} t = C n k M 1 , M 2 , ⋯ , M t M_1, M_2, \cdots, M_t M 1 , M 2 , ⋯ , M t A 1 , A 2 , ⋯ , A t A_1, A_2, \cdots, A_t A 1 , A 2 , ⋯ , A t D = M 1 A 1 + M 2 A 2 + ⋯ + M t A t D = M_1A_1 + M_2A_2 + \cdots + M_tA_t D = M 1 A 1 + M 2 A 2 + ⋯ + M t A t 0 0 0
∣ A O ∗ B ∣ = ∣ A ∗ O B ∣ = ∣ A ∣ ⋅ ∣ B ∣ \begin{vmatrix}
\mathbf{A} & \mathbf{O} \\
\mathbf{*} & \mathbf{B}
\end{vmatrix} =
\begin{vmatrix}
\mathbf{A} & \mathbf{*} \\
\mathbf{O} & \mathbf{B}
\end{vmatrix} =
\vert \mathbf{A} \vert \cdot \vert \mathbf{B} \vert A ∗ O B = A O ∗ B = ∣ A ∣ ⋅ ∣ B ∣ ∣ ∗ B A O ∣ = ∣ O B A ∗ ∣ = ( − 1 ) n m ∣ A ∣ ⋅ ∣ B ∣ \begin{vmatrix}
\mathbf{*} & \mathbf{B} \\
\mathbf{A} & \mathbf{O}
\end{vmatrix} =
\begin{vmatrix}
\mathbf{O} & \mathbf{B} \\
\mathbf{A} & \mathbf{*}
\end{vmatrix} =
(-1)^{nm} \vert \mathbf{A} \vert \cdot \vert \mathbf{B} \vert ∗ A B O = O A B ∗ = ( − 1 ) nm ∣ A ∣ ⋅ ∣ B ∣