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?
 A: 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 $\det(A)=\det(L)\det(M)\det(U)\det(V)=\det(B)\det(D)$.
A: The elementary matrix for adding a multiple of one row to another has determinant 1; thus, you can only possibly produce row operations that have determinant one. Thus, you can't perform all row operations. However, you have these building blocks:
How to swap two vectors $v$ and $w$ and negate one of them


*

*$v \leftarrow v + w$

*$w \leftarrow w - v$

*$v \leftarrow v + w$


This converts $v,w$ into $w,-v$.
This is all you need to make the row reduction algorithm work, since it only uses swaps and adding one row to another. However, you can also do
How to multiply one vector by $c$ and divide another vector by $c$:


*

*$v \leftarrow v + (1-c) w$

*$w \leftarrow w - v$

*$v \leftarrow v + \frac{c-1}{c} w$

*$w \leftarrow w + cv$


This converts $v,w$ into $(1/c)v, cw$
