$$ D = \begin{vmatrix} 1 & 2 & -4 \\ -2 & 2 & 1 \\ -3 & 4 & -2 \\ \end{vmatrix} = 1*2*(-2) + 2*1*(-3) + (-4)*(-2)*4 - 1*1*4 - 2*(-2)*(-2) - (-4)*2*(-3) = -14 $$
$$ D = \begin{vmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \\ \end{vmatrix} $$
其余子式为:
$$ M_{32}= \begin{vmatrix} a_{11} & a_{13} & a_{14} \\ a_{21} & a_{23} & a_{24} \\ a_{41} & a_{43} & a_{44} \\ \end{vmatrix} $$
代数余子式为：
$$ A_{32}= (-1)^{3+2}M_{32} = -M_{32} $$
$$ D = a_{ij}A_{ij} $$
$$ \begin{vmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\ \end{vmatrix} = \prod_{n\ge j \gt j \ge 1}(x_j - x_i) $$
$$ a_{i1}A_{j1} + a_{i2}A_{j2} + \cdots + a_{in}A_{jn}= 0, i\neq j $$
$$ D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{11} & a_{12} & \cdots & a_{1n}\\ \end{vmatrix} \neq 0 $$
唯一解为：
$$ x_1 = {D_1 \over D}, x_2 = {D_2 \over D}, \cdots, x_n = {D_n \over D} $$
其中：
$$ D_j = \begin{vmatrix} a_{11} & \cdots & a_{1.j-1} & b1 & a_{1.j+1} & \cdots & a_{1n} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ a_{n1} & \cdots & a_{n.j-1} & bn & a_{n.j+1} & \cdots & a_{nn} \\ \end{vmatrix} $$
行列式等于它的转置
$$ \begin{cases} y_1 = a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \\ y_2 = a_{21}x_1 + a_{12}x_2 + a_{23}x_3 \end{cases} $$
$$ \begin{cases} x_1 = b_{11}t_1 + b{12}t_2 \\ x_2 = b_{21}t_1 + b{22}t_2 \\ x_3 = b_{31}t_1 + b{32}t_3 \\ \end{cases} $$
线性变换得：
$$ \begin{cases} y_1 = (a_{11}b_{11} + a_{12}b_{21}+ a_{13}b_{31})t_1 + (a_{11}b_{12} + a_{12}b_{22}+ a_{13}b_{32})t_2 \\ y_2 = (a_{21}b_{11} + a_{22}b_{21}+ a_{23}b_{31})t_1 + (a_{21}b_{12} + a_{22}b_{22}+ a_{23}b_{32})t_2 \\ \end{cases} $$
矩阵的乘积：
$$ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ \end{bmatrix} \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{bmatrix} = \begin{bmatrix} a_{11}b_{11} + a_{12}b_{21}+ a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22}+ a_{13}b_{32} \\ a_{21}b_{11} + a_{22}b_{21}+ a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22}+ a_{23}b_{32} \end{bmatrix} $$
(1) (AB)C = A(BC)
(2) $\lambda(AB)C = \lambda A + \lambda B$
(3) A(B + C) = AB + AC
(4) $A^kA^l = A^{k+l}, (A^k)^l = A^{kl}$
(5) $(AB)^k \neq A^kB^k$
$$ \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \\ \end{bmatrix} ^k = \begin{bmatrix} cos k\theta & -sin k\theta \\ sin k\theta & cos k\theta \\ \end{bmatrix} $$
(1) $(A^T)^T = A$
(2) $(A + B)^T = A^T + B^T$
(3) $(\lambda A)^T = \lambda A^T$
(4) $(AB)^T = B^TA^T$
(1) $|A^T] = |A|$
(2) $|\lambda A| = \lambda^n|A|$
(3) $|AB| = |A||B|$
(4) $|AB| = |BA|。注:AB \neq BA$
$$ AA^* = A^*A = |A|E $$
对于n阶矩阵A，有一个矩阵B，使得$AB = BA = E$成立，B矩阵就是A的逆矩阵
定理1 若矩阵A可逆，则$|A|\neq 0$
定理2 若$|A|\neq 0$, 则矩阵A可逆，且
$$ A^{-1} = {1 \over |A|}A^* $$
其中$A^*是A的伴随矩阵$
|A| = 0时， A称为奇异矩阵
$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ 3 & 4 & 3 \\ \end{bmatrix} $$
解： 求得|A|=2$\neq 0, A^{-1}$存在, 计算|A|的余子式
$$ M_{11} = 2, M_{12}= 3, M_{13}= 2 $$
$$ M_{21} =-6, M_{22}=-6, M_{23}=-2 $$
$$ M_{31} =-4, M_{32}=-5, M_{33}=-2 $$
得伴随矩阵：
$$ A^* = \begin{bmatrix} M_{11} & -M_{21} & M_{31} \\ -M_{12} & M_{22} & -M_{32} \\ M_{13} & -M_{23} & M_{33} \\ \end{bmatrix} = \begin{bmatrix} 2 & 6 & -4 \\ -3 & -6 & 5 \\ 2 & 2 & -2 \\ \end{bmatrix} $$
得逆矩阵
$$ A^{-1} = {1 \over |A|}A^* = \begin{bmatrix} 1 & 3 & -2 \\ -3/2 & -3 & 5/2 \\ 1 & 1 & -1 \\ \end{bmatrix} $$
$$ A^{-1}AXBB^{-1} = A^{-1}CB^{-1} $$
得：
$$ X = A^{-1}CB^{-1} $$
$A\sim B$, 初等变换等价
$$ A = \begin{bmatrix} 3 & -1 \\ -1 & 3 \\ \end{bmatrix} $$
解：
$$ |A - \lambda E| = \begin{vmatrix} 3-\lambda & -1 \\ -1 & 3-\lambda \\ \end{vmatrix} = (3 - \lambda)^2 - 1 = (4 - \lambda)(2 - \lambda) $$
解得A的特征值为$\lambda_1 = 2, \lambda_2 = 4$
当$\lambda_1 = 2时，对应的特征向量应满足$
$$ \begin{bmatrix} 3-2 & -1 \\ -1 & 3-2 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} ，即 \begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$
解得$x_1 = x_2$, 对应的特征向量为：
$$ p_1 = \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $$
当$\lambda_2 = 4$时， 解得$x_1 = -x_2$, 对应的特征向量为：
$$ p_2 = \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix} $$