 Solution

Over a commutative field, define the determinant by the signed permutation sumIt is linear in each row separately. Swapping two rows reindexes the permutations by a transposition and reverses the sign; thus a matrix with two equal rows has determinant zero. Adding $c$ times row $i$ to a different row $j$ consequently givesThese are immediate consequences of the definition, not assumptions about elementary row operations.

For the block upper-triangular matrix, a nonzero permutation term must assign every bottom row to one of the last $n$ columns, because its first $n$ entries are zero. Those rows use all the last columns; the top rows must use all the first columns. The sign and products factor into two independent $n$-permutations, givingIn particular $X_0=\left(\begin{array}{cc}B& I\\ 0& A\end{array}\right)$ has determinant $\det B\det A$. For each bottom row $n+i$, subtract $a_{ij}$ times top row $j$ for $j=1,\dots ,n$. The top rows remain unchanged, so these additions producewithout changing the determinant. Exchange its first and last blocks of $n$ columns; the permutation has sign $(-1{)}^{n^2}$ and gives $\left(\begin{array}{cc}I& B\\ 0& -AB\end{array}\right)$. HenceSince $n^2$ and $n$ have the same parity, $\det X_1=\det (AB)$. Combining the two evaluations provesNo invertibility of $A$ or $B$ was used, so the proof includes singular matrices.