性质 4.1 转置保持行列式不变

性质 4.1. 转置保持行列式不变

行与列互换, 行列式不变, 即 $\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|=\left|\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_{n1}\\ a_{12}& a_{22}& \cdots & a_{n2}\\ ⋮& ⋮& & ⋮\\ a_{1n}& a_{2n}& \cdots & a_{nn}\end{array}\right|.$

\begin{proof} 设 $b_{ij}=a_{ji}\left(i,j=1,2,\cdots ,n\right)$ ,则

$$\begin{array}{rl}\text{ 右边 }& =\left|\begin{array}{cccc}{b}_{11}& b_{12}& \cdots & b_{1n}\\ b_{21}& b_{22}& \cdots & b_{2n}\\ ⋮& ⋮& & ⋮\\ b_{n1}& b_{n2}& \cdots & b_{nn}\end{array}\right|\\ & =\sum _{j_1{j}_2\cdots j_n}{\left(-1\right)}^{\tau \left(j_1{j}_2\cdots j_n\right)}{b}_{j_{1}1}{b}_{j_{2}2}\cdots b_{j_{n}n}\\ & =\sum _{j_1{j}_2\cdots j_n}{\left(-1\right)}^{\tau \left(j_1{j}_2\cdots j_n\right)}{a}_{1j_1}{a}_{2j_2}\cdots a_{n{j}_n}=\text{ 左边. }\end{array}$$

\end{proof}

假设 $D=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|,D^{\mathrm{T}}=\left|\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_{n1}\\ a_{12}& a_{22}& \cdots & a_{n2}\\ ⋮& ⋮& & ⋮\\ a_{1n}& a_{2n}& \cdots & a_{nn}\end{array}\right|,$ 那么称 $D^{\mathrm{T}}$ 为 $D$ 的转置行列式 (transpose), 有时候也记作 $D^{\prime }$. 于是, 性质 4.1 可以简写为 $D=D^{\mathrm{T}}.$ 由此性质可知行列式中行和列的地位是对称的,具有相同的性质.