<s>[INST] Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12}  & \cdots & a_{1k} &0  &\cdots   &0 \\ 
 a_{21}&  a_{22}& \cdots  & a_{2k} & 0  &\cdots   &0 \\ 
 \vdots& \vdots &  & \vdots & \vdots &  &\vdots\\ 
a_{k1} & a_{k2} & \cdots  & a_{kk} & 0 &\cdots & 0\\ 
 c_{11}& c_{12} &  \cdots& c_{1k} & b_{11} & \cdots  & b_{1m}\\ 
\vdots& \vdots &  & \vdots &  b_{21}&\cdots   & b_{2m}\\ 
 c_{m1}& c_{m2} & \cdots  & c_{mk} &  b_{m1}& \cdots  & b_{mm}
\end{vmatrix} 
\end{align}
If I have got a determinant $$D_1=
\begin{vmatrix}
 0  &\cdots   &0&a_{11} &a_{12}  & \cdots & a_{1k} \\ 
 0  &\cdots   &0 &a_{21}&  a_{22}& \cdots  & a_{2k}  \\ 
 \vdots&  & \vdots & \vdots & \vdots &  &\vdots\\ 
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots  & a_{kk} \\ 
 b_{11} & \cdots  & b_{1m}& c_{11}& c_{12} &  \cdots& c_{1k}\\ 
\vdots&  & \vdots & \vdots &  \vdots&   & \vdots\\ 
  b_{m1}& \cdots  & b_{mm}&c_{m1}& c_{m2} & \cdots  & c_{mk} 
\end{vmatrix} 
$$
Then $D_1$ is equal to $(-1)^{k \times m}$$D$.
I know that the existence of the factor -1 is due to the interchange of 2 row, but i have a question on that $k \times m$.In my book,it said that i have to do $k\times m$ times row operations to transform $D_1$ into $D$.However,i thought only k times is needed for $D_1$ transform into $D$.
If i have done 1 times row operation for $D_1$
$$D_1=
\begin{vmatrix}
 0  &\cdots   &0&a_{11} &a_{12}  & \cdots & a_{1k} \\ 
 0  &\cdots   &0 &a_{21}&  a_{22}& \cdots  & a_{2k}  \\ 
 \vdots&  & \vdots & \vdots & \vdots &  &\vdots\\ 
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots  & a_{kk} \\ 
 b_{11} & \cdots  & b_{1m}& c_{11}& c_{12} &  \cdots& c_{1k}\\ 
\vdots&  & \vdots & \vdots &  \vdots&   & \vdots\\ 
  b_{m1}& \cdots  & b_{mm}&c_{m1}& c_{m2} & \cdots  & c_{mk} 
\end{vmatrix} 
=\begin{vmatrix}
 a_{11}  &\cdots   &0&0 &a_{12}  & \cdots & a_{1k} \\ 
 a_{21} &\cdots   &0 &0&  a_{22}& \cdots  & a_{2k}  \\ 
 \vdots&  & \vdots & \vdots & \vdots &  &\vdots\\ 
a_{k1} &\cdots & 0& 0& a_{k2} & \cdots  & a_{kk} \\ 
 c_{11} & \cdots  & b_{1m}&b_{11} & c_{12} &  \cdots& c_{1k}\\ 
\vdots&  & \vdots & \vdots &  \vdots&   & \vdots\\ 
  c_{m1}& \cdots  & b_{mm}&b_{m1}& c_{m2} & \cdots  & c_{mk} 
\end{vmatrix} 
$$
Correct me if i have made any mistakes
 [/INST] You can think of it in this way: take column $k+1$ (the first one containing $b$s) and swap it with column $k$. Then swap column $k$ with column $k-1$, etc, until you have "bubbled" the original column $k+1$ all the way to the left, after a total of $k$ swaps.
Repeating this process for all $m$ columns on the right requires a total of $km$ swaps.
</s>