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.

• Some typesetting advices: bmatrix environment is better suited for matrices that array (you get nicer code, and need not "predict" the number of columns). Also, align* environment will help you make nice multiline formulas, so they don't "run out" of the post. – Vedran Šego Aug 26 '13 at 10:26
• It appears that you could let $X=0$ from beginning to simplify proof. – Maesumi Aug 26 '13 at 11:09
• See reference to block matrices here – Maesumi Aug 26 '13 at 11:45

$$A = \begin{bmatrix} 2 & 0 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 \end{bmatrix}.$$
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).