Determinant of $2\times 2$ block matrices I am trying to solve the problem here: Let $A,B,C,D,$ be commuting $n\times n$ matrices over the field $F$ (it is not given whether any of these matrices are invertible). Show that the determinant of the $2n\times 2n$ matrix
$$\begin{bmatrix}
A&B\\C&D
\end{bmatrix}$$
is $\det(AD-BC)$.
I know that there is an answer given in a paper, which mentions working over $F[x]$. However, I have come up with the following idea:

Motivated by the adjoint formula for the inverse, we get that
  $$\begin{bmatrix}
A&B\\ C&D
\end{bmatrix}
\begin{bmatrix}
D&-B\\ -C&A
\end{bmatrix}=
\begin{bmatrix}
AD-BC & BA-AB\\ CD-DC&AD-BC
\end{bmatrix}
=
\begin{bmatrix}
AD-BC & 0\\ 0&AD-BC
\end{bmatrix}
$$
  Hence
  $$\det \left(\begin{bmatrix}
A&B\\ C&D
\end{bmatrix}\right)\det\left(\begin{bmatrix}
D&-B\\ -C&A
\end{bmatrix}\right)= \det(AD-BC)^2$$
  But then 
  $$\det\left( \begin{bmatrix}
A&B\\ C&D
\end{bmatrix}\right)=
-\det\left(\begin{bmatrix}
B&A\\ D&C
\end{bmatrix}\right)=
-\det\left(\begin{bmatrix}
B&D\\ A&C
\end{bmatrix}\right)=
\det\left(\begin{bmatrix}
D&B\\ C&A
\end{bmatrix}\right)
$$
  Now, thinking of $\det$ in terms of permutations, suppose for a given permutation in the $\det$ sum that we pick $m$ elements in the first $n$ rows from the $-B$ side. Then we must also pick $m$ elements from the last $n$ rows from the $-C$ side; i.e. the sign changes must cancel out. Thus,
  $$\det\left(\begin{bmatrix}
D&B\\ C&A
\end{bmatrix}\right)=\det\left(\begin{bmatrix}
D&-B\\ -C&A
\end{bmatrix}\right)$$
  which shows that
  $$\det \left(\begin{bmatrix}
A&B\\ C&D
\end{bmatrix}\right)=\pm \det(AD-BC)$$

Is there a way complete the proof from here? All I need to show is that
$$\det \left(\begin{bmatrix}
A&B\\ C&D
\end{bmatrix}\right)\quad\text{ and }\quad \det(AD-BC)$$
have the same sign.
 A: $\textbf{Remarks.}$
i) After "but then", that is correct $$\det\left( \begin{bmatrix}
A&B\\ C&D
\end{bmatrix}\right)=
(-1)^n\det\left(\begin{bmatrix}
B&A\\ D&C
\end{bmatrix}\right)=
\det\left(\begin{bmatrix}
D&C\\ B&A
\end{bmatrix}\right)
$$
ii) Why do you write 
$$
\det\left(\begin{bmatrix}
B&A\\ D&C
\end{bmatrix}\right)=
\det\left(\begin{bmatrix}
B&D\\ A&C
\end{bmatrix}\right)\;?
$$ 
There is no transposition here.
iii) Last line. What do you mean by "same signum " in a field $F$.
EDIT. That follows is a simple proof  -using the polynomials-
Let $\overline{F}$ be the algebraic closure of $F$. It is known that  $\overline{F}$ is an infinite field.
We know also that the required formula is valid when $A,B,C,D\in M_n(\overline{F})$ and $A$ is invertible; if not, we replace $A$ with $A+\lambda I_n$ where $\lambda\in \overline{F}$ is s.t. $P(\lambda)=\det(A+\lambda I_n)\not= 0$; note that, since $P$ admits at most $n$ roots, the set $S$ of such $\lambda$'s is infinite. 
Then, for every $\lambda\in S$,  $$Q(\lambda)=
\det\left(\begin{bmatrix}
A+\lambda I_n&B\\ C&D
\end{bmatrix}\right)-\det(AD-BC+\lambda D)=0.$$
The polynomial $Q$ admits an infinity of roots, and therefore, $Q(0)=0$.  $\square$
