$\textbf{Problem}$: Let a $2n \times 2n$ matrix be given in the form $M=\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]$, where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$. Use block multiplication to prove that $\det M= \det(AD-CB)$. Give an example to show that this formula need not hold if $AC \neq CA$
$\textbf{Proof}$: Let $A,B,C,D,X \in \textbf{M}_n(K)$ such that $A+BX$ is invertible.
For all $Y \in \textbf{M}_n(K)$, we have:
$$\left[ {\begin{array}{cc}
I_n & 0 \\
Y & I_n \\
\end{array} } \right] \left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right] \left[ {\begin{array}{cc}
I_n & 0 \\
X & I_n \\
\end{array} } \right]= \left[ {\begin{array}{cc}
A+BX & B \\
YA+C+(YB+D)X & YB+D \\
\end{array} } \right].$$
Let $Y=-(C+DX)(A+BX)^{-1}$. Hence:
$$YA+C+(YB+D)X=Y(A+BX)+(C+DX)=0.$$
Since $\det\left[ {\begin{array}{cc}
I_n & 0 \\
Y & I_n \\
\end{array} } \right]= \det\left[ {\begin{array}{cc}
I_n & 0 \\
X & I_n \\
\end{array} } \right]= (\det(I_n))^2=1$, we can conclude that:
\begin{align*}
\det\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]&=\det\left[ {\begin{array}{cc}
A+BX & B \\
0 & YB+D \\
\end{array} } \right]\\
&= \det(A+BX)\det(-(C+DX)(A+BX)^{-1}B+D).
\end{align*}
In particular for $X=0$, we have:
\begin{align*}
\det\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]&=\det(A)\det(-CA^{-1}B+D)=\det(-ACA^{-1}B+AD) \\
&=\det(-CAA^{-1}B+AD)=\det(AD-CB).
\end{align*}
I just wanted someone to verify my proof and help me with the second part of this question.
Thank you in advance
Best Answer
Your proof seems fine to me.
As for a counterexample, consider
$$A = \begin{bmatrix} 2 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 \end{bmatrix}.$$
In one hand, $\det A = -4$ (check here), and in the other hand, $\det (A_{11} A_{22} - A_{21}A_{12}) = 0$ (check here).