Determinant of matrix obtained by commuting matrices The Question is to prove that :
For Commuting $n\times n$ matrices $A,B,C,D$ over a field $F$,
Determinant of $\left(\begin{array}{cccc}
A & B  \\
C & D  \\
\end{array} \right)$ is given by $\det(AD-BC)$
I have no idea how to proceed for this except at the case of $n=1$ where $\left(\begin{array}{cccc}
A & B  \\
C & D  \\
\end{array} \right)$=$\begin{pmatrix}
a& b  \\
c & d  \\
\end{pmatrix}$ for some $a,b,c,d\in F$ and I know $\det\left(\begin{array}{cccc}
a& b  \\
c & d  \\
\end{array} \right)=ad-bc=\det(ad-bc)=\det(AD-BC)$
So, for $n=1$ we have $\det\left(\begin{array}{cccc}
A & B  \\
C & D  \\
\end{array} \right)=\det(AD-BC)$
I have no idea how to proceed for general $n$ not even when $n=2$
Do I need to proceed by induction?I doubt that it may not work..
please provide some hints to prove this case...
Thank You.
 A: This is a partial duplicate of the thread "block matrices problem". You question is addressed in the last paragraph of my answer.
More importantly, note that if only a pair of matrices (but not four of them) on the same row or same column commute, the order of matrices matters. In short, we have
$$
\det\pmatrix{A&B\\ C&D}=
\begin{cases}
\det(AD-BC) & \text{ if } CD=DC,\\
\det(DA-CB) & \text{ if } AB=BA,\\
\det(DA-BC) & \text{ if } BD=DB,\\
\det(AD-CB) & \text{ if } AC=CA.
\end{cases}
$$
A: Here is a method that works if $F = \mathbb{C}$. Let $\mathcal{F} = \{A, B, C, D\}$ and let $P\in M_{n}(\mathbb{C})$ such that $P^{-1}XP$ is upper triangular for each $X\in \mathcal{F}$. Then
\begin{equation*}
\det\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} = \det\begin{pmatrix}
P^{-1} & 0\\
0 & P^{-1}
\end{pmatrix}\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}\begin{pmatrix}
P & 0\\
0 & P
\end{pmatrix} = \det\begin{pmatrix}
P^{-1}AP & P^{-1}BP\\
P^{-1}CP & P^{-1}DP
\end{pmatrix}.
\end{equation*}
Further, $\det(AD-BC) = \det(P^{-1}(AD-BC)P) = \det((P^{-1}AP)(P^{-1}DP)-(P^{-1}BP)(P^{-1}CP))$. Thus it suffices to assume that $A$, $B$, $C$ and $D$ are all upper triangular. We prove the result by induction. For $n = 1$ there is nothing to prove. Expanding the determinant about the first column we have
\begin{align*}
    \det\left(\begin{array}{ccc|ccc}
  a_{11} &  & \ast & b_{11} &  & \ast\\
   & \ddots &  &  & \ddots & \\
   &  & a_{nn} &  &  & b_{nn}\\
  \hline
  c_{11} &  & \ast & d_{11} &  & \ast\\
   & \ddots &  &  & \ddots & \\
   &  & c_{nn} &  &  & d_{nn}
\end{array}\right) =& (-1)^{1+1}a_{11}\det\left(\begin{array}{ccc|cccc}
  a_{22} &  & \ast & 0 & b_{22} &  & \ast\\
   & \ddots &  & \vdots &  & \ddots & \\
   &  & a_{nn} & 0 &  &  & b_{nn}\\
  \hline
  c_{12} & \dots & c_{1n} & d_{11} &  &  & \ast\\
  c_{22} &  & \ast &  & \ddots &  &  \\
   & \ddots &  &  &  & \ddots & \\
   &  & c_{nn} &  &  &  & d_{nn}
\end{array}\right)\\
&+(-1)^{1+n+1}c_{11}\det\left(\begin{array}{ccc|cccc}
  a_{12} & \dots & a_{1n} & b_{11} &  &  & \ast\\
  a_{22} &  & \ast &  & \ddots &  & \\
   & \ddots &  &  &  & \ddots & \\
   &  & a_{nn} &  &  &  & b_{nn}\\
  \hline
  c_{22} &  & \ast & 0 & d_{22} &  & \ast\\
   & \ddots &  & \vdots &  & \ddots & \\
   &  & c_{nn} & 0 &  &  & d_{nn}
\end{array}\right).
\end{align*}
Expanding both terms about the $n^{\text{th}}$ column we get
\begin{equation*}
    \det\left(\begin{array}{c|c}
        A & B\\
         \hline
        C & D
    \end{array}\right) = (a_{11}d_{11}-b_{11}c_{11})\det\left(\begin{array}{ccc|ccc}
  a_{22} &  & \ast & b_{22} &  & \ast\\
   & \ddots &  &  & \ddots & \\
   &  & a_{nn} &  &  & b_{nn}\\
  \hline
  c_{22} &  & \ast & d_{22} &  & \ast\\
   & \ddots &  &  & \ddots & \\
   &  & c_{nn} &  &  & d_{nn}
\end{array}\right).
\end{equation*}
Conclude using the inductive hypothesis.
