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.
Best Answer
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 $$ \det\pmatrix{A&B\\ C&D}= \begin{cases} \det(AD-BC) & \text{ if } CD=DC,\\ \det(DA-CB) & \text{ if } AB=BA,\\ \det(DA-BC) & \text{ if } BD=DB,\\ \det(AD-CB) & \text{ if } AC=CA. \end{cases} $$