= Solution
Write the <determinant> in its <Leibniz formula for determinants>, where $S_n$ is the set of <permutations> of $\{1,\ldots,n\}$:
$$
\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}.
$$
The <permutation sign> is justified by the given row-interchange property: a <permutation matrix> is obtained from the <identity matrix> by row interchanges, and every interchange changes its <determinant> by a factor $-1$. Reversing a sequence of interchanges constructs the inverse <permutation> with the same number of interchanges. Consequently $\operatorname{sgn}(\sigma^{-1})=\operatorname{sgn}(\sigma)$. The <Leibniz formula for determinants> can itself be obtained by expanding the rows multilinearly in the <standard basis>: repeated selected coordinate rows contribute zero, and the remaining terms are precisely these signed <permutation matrices>.
Since $(A^T)_{ij}=a_{ji}$, the same formula gives
$$
\det(A^T)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{\sigma(i),i}.
$$
Reindex the product with $j=\sigma(i)$ and reindex the sum with $\tau=\sigma^{-1}$. Both reindexings are bijections, so
$$
\det(A^T)=\sum_{\tau\in S_n}\operatorname{sgn}(\tau)\prod_{j=1}^n a_{j,\tau(j)}=\det(A).
$$
Thus \b[$\det(A)=\det(A^T)$] for every real square <matrix>, including <singular matrices>. This proof does not assume the analogous column-operation rules in order to establish the <transpose> identity.
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