Solution
= Solution
The <Leibniz formula for determinants> gives $\det(A^T)=\sum_{\sigma\in S_3}\operatorname{sgn}(\sigma)\prod_i a_{\sigma(i),i}$. Relabel the product by $j=\sigma(i)$ and then the sum by $\tau=\sigma^{-1}$. The inverse <permutation> has the same sign, so
$$
\det(A^T)=\sum_{\tau\in S_3}\operatorname{sgn}(\tau)\prod_j a_{j,\tau(j)}=\det A.
$$
The proof applies to any finite square <matrix>.