Is it possible to triangularize a matrix only by adding scalar multiples of rows to each other? I am working on showing if $B$ is a $s \times s$ matrix, $D$ is a $t \times t$  matrix, $C$ is a $s \times t$ matrix, and $0$ is a $t \times s$ zero matrix, then $\det(A)=\det(B)\det(D)$, where $$A = \begin{bmatrix}B &C\\ 0& D\end{bmatrix}$$
I understand two more complicated proofs, but I figured out if I could triangularize a matrix by just adding scalar multiples of row i to row j (in that sense my determinant will not change), then I can make to proof much more easier.
However, is that generally true?
After put in some thought, I realized I can use all elementary row operations in my proof. It will not affect the proof. A scratch of my proof will be I am trying to triangularize B and D and A as well, then because the determinant is just the product of diagonal element I can show $\det(A)=\det(B)\det(D)$. If I used row swaps and multiply a row by a scalar, I will have shown $K\det(A)=K\det(B)\det(D)$, where K is some constant due to the row operations. This then gives the result I want.
So, the question no longer have anything to do with the proof now. But still, is it possible?

Use the LU decompositions $B=LU,D=MV$ where $L,M$ are lower triangular with diagonal $1$, then invertible and $U,V$ are upper triangular (we assume that there are no permutation matrices). Thus $A=\begin{pmatrix}L&0\\0&M\end{pmatrix}\begin{pmatrix}U&L^{-1}C\\0&V\end{pmatrix}$ and $\det(A)=\det(L)\det(M)\det(U)\det(V)=\det(B)\det(D)$.
