Question in the proof of the formula of the determinant of a block lower triangular matrix in Huffman - Kunze Huffman and Kunze first show that if $B$ is obtained from $A$ by adding the rows $c\alpha_j$ to $\alpha_i$ where $i<j$ then $\det A=\det B$. Then they define $D(A,B,C)=\det\begin{bmatrix}
A & B \\ 0 & C
\end{bmatrix}$, if $A$ and $B$ are fixed then $D$ is alternating and $s$-linear as a function of the rows of $C$, so by a previous theorem $$D(A,B,C)=(\det C) D(A,B,I)$$ where $I$ is the identity function with $s\times s$ dimension. Now they claim that: 

By substracting multiples of the rows of $I$ from the rows of $B$ and using (5-18) [i.e. the first $\det A=\det B$ result above when $B$ is obtained from $A$ by adding a linear combination of its rows to another] we obtain: $$D(A,B,I)=D(A,0,I)$$

But I don't know how that was exactly obtained, remembering that $B$ has $r\times s$ dimensions, so how could we ensure that we have enough rows in $I_{s\times s}$ to cover $r$ entirely and why is it equal in the first place.
 A: Let $$P=\pmatrix{A&B\\ 0&I}$$ and $$Q=\pmatrix{A&0\\ 0&I}.$$ Let $\pmatrix{a_{11}&\cdots&a_{1r}&b_{11}&\cdots&b_{1s}}$ the first row of $P$. Subtract $b_{11}\times(\text{$(r+1)^{{\rm th}}$ row of $P$})$ from the first row. This gives us $\pmatrix{a_{11}&\cdots&a_{1r}&0&b_{12}&\cdots&b_{1s}}$.  Denote the new matrix obtained by $P_1$. By $(5.18)$, we have $\det P=\det P_1$.
Now subtract $b_{12}\times(\text{$(r+2)^{{\rm th}}$ row of $P_1$})$ from the first row $P_1$. This gives us a new matrix $P_2$ with first row as $\pmatrix{a_{11}&\cdots&a_{1r}&0&0&b_{13}&\cdots&b_{1s}}$. Also, note that $\det P=\det P_1=\det P_2$. Since there are $(r+s)$ rows available, by repeatdly applying the above procedure, one can obtain a matrix $R$ with first row as $\pmatrix{a_{11}&\cdots&a_{1r}&0&0&0&\cdots&0}$ and $\det P=\det R$. 
Do the same thing for other rows and obtain the matrix $Q$ with $\det P=\det Q$. Since  by definition, $D(A,B,I)=\det P$ and $D(A,0,I)=\det Q$, we obtain the required result.
A: Note that
$$
\pmatrix{A&0\\ 0&I}=\pmatrix{I&-B\\ 0&I}\pmatrix{A&B\\ 0&I}.
$$
Since $\pmatrix{I&-B\\ 0&I}=\prod_{i=1}^r\prod_{j=1}^s\pmatrix{I&-b_{ij}E_{ij}\\ 0&I}$ is a product of elementary matrices (where $E_{ij}$ denotes the $r\times s$ matrix whose only nonzero entry is a $1$ at the $(i,j)$-th position), it represents a series of row operations.
