Es sei ${\displaystyle {}M={\left(a_{ij}\right)}}$ und ${\displaystyle {}{M^{\text{tr}}}={\left(b_{ij}\right)}}$ mit ${\displaystyle {}b_{ij}=a_{ji}}$. Dann ist

${\displaystyle {}{\begin{aligned}\delta {\left({M^{\text{tr}}}\right)}&=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}b_{i\pi (i)}\\&=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )\prod _{i=1}^{n}a_{\pi (i)i}\\&=\sum _{\rho =\pi ^{-1}\in S_{n}}\operatorname {sgn} (\rho ^{-1})\prod _{i=1}^{n}a_{\rho ^{-1}(i)i}\\&=\sum _{\rho \in S_{n}}\operatorname {sgn} (\rho )\prod _{j=1}^{n}a_{j\rho (j)}\\&=\delta (M).\end{aligned}}}$