这是《玩转线性代数》的学习笔记 少壮不努力，老大徒伤悲，学校里没学好，工作多年后又从头看，两行泪。。。

将符号 ∣ a 11 a 12 a 21 a 22 ∣ \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix} ∣∣∣∣​a11​a21​​a12​a22​​∣∣∣∣​称为二阶行列式，它的值为 a 11 a 22 − a 12 a 21 a_{11}a_{22}-a_{12}a_{21} a11​a22​−a12​a21​，记为 D D D，即 D = ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 a 21 D=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21} D=∣∣∣∣​a11​a21​​a12​a22​​∣∣∣∣​=a11​a22​−a12​a21​

用主对角线的乘积减去副对角线乘积的方式计算行列式的方法称为对角线法则

解二元线性方程组： { a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 （ 2.1 ） \left\{ \begin{aligned} a_{11}x_1+a_{12}x_2 = b_1\\ a_{21}x_1+a_{22}x_2 = b_2 \end{aligned} \right. \quad（2.1） {a11​x1​+a12​x2​=b1​a21​x1​+a22​x2​=b2​​（2.1） 解得解为 { x 1 = b 1 a 22 − b 2 a 12 a 11 a 22 − a 12 a 21 x 2 = b 2 a 11 − b 1 a 21 a 11 a 22 − a 12 a 21 （ a 11 a 22 − a 12 a 21 ≠ 0 ） \left\{ \begin{aligned} x_1 = \frac{b_{1}a_{22}-b_{2}a_{12}}{a_{11}a_{22}-a_{12}a_{21}}\\ x_2 = \frac{b_{2}a_{11}-b_{1}a_{21}}{a_{11}a_{22}-a_{12}a_{21}}\\ \end{aligned} \right. \quad（a_{11}a_{22}-a_{12}a_{21}\neq0） ⎩⎪⎪⎨⎪⎪⎧​x1​=a11​a22​−a12​a21​b1​a22​−b2​a12​​x2​=a11​a22​−a12​a21​b2​a11​−b1​a21​​​（a11​a22​−a12​a21​​=0） 记 D 1 = ∣ b 1 a 12 b 2 a 22 ∣ = b 1 a 22 − b 2 a 12 D 2 = ∣ a 11 b 1 a 21 b 2 ∣ = a 11 b 2 − b 1 a 21 D_1 = \begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22}\end{vmatrix}=b_{1}a_{22}-b_{2}a_{12}\\ D_2 = \begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2}\end{vmatrix}=a_{11}b_{2}-b_{1}a_{21} D1​=∣∣∣∣​b1​b2​​a12​a22​​∣∣∣∣​=b1​a22​−b2​a12​D2​=∣∣∣∣​a11​a21​​b1​b2​​∣∣∣∣​=a11​b2​−b1​a21​ 则当 D = ∣ a 11 a 12 a 21 a 22 ∣ ≠ 0 D=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}\neq0 D=∣∣∣∣​a11​a21​​a12​a22​​∣∣∣∣​​=0时，方程组的解惟一且可以表示为： x 1 = D 1 D , x 2 = D 2 D x_1=\frac{D_1}{D},x_2=\frac{D_2}{D} x1​=DD1​​,x2​=DD2​​

将符号 ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ \begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix} ∣∣∣∣∣∣​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​∣∣∣∣∣∣​称为三阶行列式，它的值为 a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 − a 13 a 22 a 31 − a 11 a 23 a 32 − a 12 a 21 a 33 a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31}+ a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33} a11​a22​a33​+a12​a23​a31​+a13​a21​a32​−a13​a22​a31​−a11​a23​a32​−a12​a21​a33​

将元素循环移位，可以看到同样满足对角线法则 图来自原地址

解三元线性方程组： { a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 \left\{ \begin{aligned} a_{11}x_1+a_{12}x_2+a_{13}x_3 = b_1\\ a_{21}x_1+a_{22}x_2+a_{23}x_3 = b_2\\ a_{31}x_1+a_{32}x_2+a_{33}x_3 = b_3\\ \end{aligned} \right. \quad ⎩⎪⎨⎪⎧​a11​x1​+a12​x2​+a13​x3​=b1​a21​x1​+a22​x2​+a23​x3​=b2​a31​x1​+a32​x2​+a33​x3​=b3​​ 令 D = ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ ≠ 0 , D 1 = ∣ b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 ∣ , D 2 = ∣ a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 ∣ , D 3 = ∣ a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 ∣ ( D 1 , D 2 , D 3 的 意 义 与 上 同 ) D=\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}\neq0, D_1=\begin{vmatrix} b_{1} & a_{12} & a_{13}\\ b_{2} & a_{22} & a_{23} \\ b_{3} & a_{32} & a_{33}\end{vmatrix}, \\ D_2=\begin{vmatrix} a_{11} & b_{1} & a_{13}\\ a_{21} & b_{2} & a_{23} \\ a_{31} & b_{3} & a_{33}\end{vmatrix}, D_3 =\begin{vmatrix} a_{11} & a_{12} & b_{1}\\ a_{21} & a_{22} & b_{2} \\ a_{31} & a_{32} & b_{3}\end{vmatrix} \quad (D_1,D_2,D_3的意义与上同) D=∣∣∣∣∣∣​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​∣∣∣∣∣∣​​=0,D1​=∣∣∣∣∣∣​b1​b2​b3​​a12​a22​a32​​a13​a23​a33​​∣∣∣∣∣∣​,D2​=∣∣∣∣∣∣​a11​a21​a31​​b1​b2​b3​​a13​a23​a33​​∣∣∣∣∣∣​,D3​=∣∣∣∣∣∣​a11​a21​a31​​a12​a22​a32​​b1​b2​b3​​∣∣∣∣∣∣​(D1​,D2​,D3​的意义与上同)

若 D ≠ 0 D\neq0 D​=0，则有唯一解： x 1 = D 1 D , x 2 = D 2 D , x 3 = D 3 D x_1=\frac{D_1}{D},x_2=\frac{D_2}{D},x_3=\frac{D_3}{D} x1​=DD1​​,x2​=DD2​​,x3​=DD3​​