# 矩阵乘法中的矩阵不满足消去律和交换律，但矩阵对应的行列式满足消去律和交换律

## 二、解析

### 矩阵乘法不满足消去律

$$| \boldsymbol { A } \boldsymbol { B } | = | \boldsymbol { A } | | \boldsymbol { B } | = 0$$

1. $| \boldsymbol { A } |$ $=$ $0$ 且 $| \boldsymbol { B } |$ $=$ $0$
2. $| \boldsymbol { A } |$ $=$ $0$ 且 $| \boldsymbol { B } |$ $\neq$ $0$
3. $| \boldsymbol { A } |$ $\neq$ $0$ 且 $| \boldsymbol { B } |$ $=$ $0$

$$\textcolor{orangered}{\boldsymbol { A B } = \boldsymbol { O } \nRightarrow \boldsymbol { A } = \boldsymbol { O } }$$

$$\textcolor{orangered}{\boldsymbol { A B } = \boldsymbol { O } \nRightarrow \boldsymbol { B } = \boldsymbol { O } }$$

### 矩阵乘法不满足交换律

$$\textcolor{orangered}{\boldsymbol { A B } \neq \boldsymbol { B A } }$$

### 例子

\begin{aligned} \boldsymbol {A B} \\ \\ & = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ – 1 & – 1 \end{pmatrix} \\ \\ & = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \\ \\ & = \boldsymbol{O} \end{aligned}

$$\boldsymbol { A } \neq \boldsymbol { O }$$

$$\boldsymbol { B } \neq \boldsymbol { O }$$

\begin{aligned} \boldsymbol { B } \boldsymbol { A } \\ \\ & = \begin{pmatrix} 1 & 1 \\ – 1 & – 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \\ \\ & = \begin{pmatrix} 2 & 2 \\ -2 & -2 \end{pmatrix} \\ \\ & \neq \boldsymbol { O } \\ \\ & \neq \boldsymbol{ AB } \end{aligned}