<s>[INST] Determinant of a $2 \times 2$ block matrix $\textbf{Problem}$: 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
 [/INST] 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).
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