# 如何确定行列式展开计算公式中每一项的正负？

## 一、前言

$$\left|\begin{matrix} a_{11} & a_{12} & \cdots & a_{1⁢⁢⁢n} \\ a_{21} & a_{22} & \cdots & a_{2⁢⁢⁢n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{m}\end{matrix}\right| = \textcolor{yellow}{\sum _{j_{1} j_{2} \cdots j_{n}}} \textcolor{springgreen}{\left(−1\right)^{\tau \left(j_{1}j_{2} \cdots j_{n}\right)}} \textcolor{pink}{a_{1j_{1}}a_{2j_{2}} \cdots a_{n}}$$

## 二、正文

$$a_{1 j_{1}} \cdot a_{2 j_{2}} \cdot a_{3 j_{3}} \cdots a_{n j_{n}}$$

$$\tau \left( j_{1} j_{2} \cdots j_{n} \right)$$

$$123 \cdots n$$

$$\textcolor{yellow}{\tau \left( j_{1} j_{2} \cdots j_{n} \right)} + \textcolor{springgreen}{\boldsymbol{0}} = \textcolor{yellow}{\tau \left( j_{1} j_{2} \cdots j_{n} \right)}$$

\begin{aligned} (-1) ^{\text{偶数}} & = 1 \\ (-1) ^{\text{奇数}} & = -1 \end{aligned}

\begin{aligned} \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix} & = \\ \\ & + \left( a_{11} \cdot a_{22} \cdot a_{33} \right) \\ & \textcolor{springgreen}{+ \left( a_{12} \cdot a_{23} \cdot a_{31} \right)} \\ & + \left( a_{13} \cdot a_{32} \cdot a_{21} \right) \\ & – \left( a_{13} \cdot a_{22} \cdot a_{31} \right) \\ & \textcolor{orangered}{- \left( a_{12} \cdot a_{21} \cdot a_{33} \right)} \\ & – \left( a_{11} \cdot a_{32} \cdot a_{23} \right) \end{aligned}

$$\begin{cases} \tau \left( \textcolor{pink}{123} \right) = \textcolor{pink}{0} \\ \tau \left( \textcolor{red}{231} \right) = \textcolor{red}{2} \end{cases}$$

$$\left( -1 \right) ^{\textcolor{pink}{0} + \textcolor{red}{2} } = 1$$

$$\begin{cases} \tau \left( \textcolor{pink}{213} \right) = \textcolor{pink}{1} \\ \tau \left( \textcolor{red}{321} \right) = \textcolor{red}{3} \end{cases} \Rightarrow \left( -1 \right) ^{\textcolor{pink}{1} + \textcolor{red}{3} } = 1$$

$$\begin{cases} \tau \left( \textcolor{springgreen}{123} \right) = \textcolor{springgreen}{0} \\ \tau \left( \textcolor{yellow}{213} \right) = \textcolor{yellow}{1} \end{cases} \Rightarrow \left( -1 \right) ^{\textcolor{springgreen}{0} + \textcolor{yellow}{1} } = -1$$

$$\begin{cases} \tau \left( \textcolor{springgreen}{321} \right) = \textcolor{springgreen}{3} \\ \tau \left( \textcolor{yellow}{312} \right) = \textcolor{yellow}{2} \end{cases} \Rightarrow \left( -1 \right) ^{\textcolor{springgreen}{3} + \textcolor{yellow}{2} } = -1$$

$$-a_{13} \cdot a_{2 \textcolor{orangered}{i}} \cdot a_{46} \cdot a_{34} \cdot a_{61} \cdot a_{5 \textcolor{orangered}{j} }$$

$$\begin{cases} i = ? \\ j = ? \end{cases}$$

$$\tau \left( 124365 \right) = 2$$

$$3, \textcolor{orangered}{i}, 6, 4, 1, \textcolor{orangered}{j}$$

$$2, 5$$

$$\tau \left( 356412 \right) = 10$$

$$\tau \left( 326415 \right) = 7$$

$$\begin{cases} i = 2 \\ j = 5 \end{cases}$$