Addition-related property of the determinant of a $2 \times 2$ block matrix

Question

For $2 \times 2$ matrices,

$$\det \begin{bmatrix} a&b \\ c+e&d \end{bmatrix} =\det\begin{bmatrix} a&b \\ c&d \end{bmatrix}+ \det \begin{bmatrix} a&b \\ e&0 \end{bmatrix}$$

If $A$ is $m \times m$ matrix and $D$ is $n\times n$ matrix, is the following correct?

$$\det \begin{bmatrix} A&B \\ C+E&D \end{bmatrix} =\det\begin{bmatrix} A&B \\ C&D \end{bmatrix}+ \det \begin{bmatrix} A&B \\ E&0 \end{bmatrix}$$

If it is correct, please show me a proof.

Answer 1

In general, no. Suppose that $A$, $B$, $C$, and $D$ are $2\times2$ matrices, with $A=B=D=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$. Take $C=\left[\begin{smallmatrix}c_1&c_2\\c_3&c_4\end{smallmatrix}\right]$ and $E=\left[\begin{smallmatrix}e_1&e_2\\e_3&e_4\end{smallmatrix}\right]$. Then\begin{multline}\det\begin{bmatrix}A&B\\C+E&D\end{bmatrix}=\\=1 - c_1 - c_2 c_3 - c_4 + c_1 c_4 - e_1 + c_4 e_1 - c_3 e_2 - c_2 e_3 - e_2 e_3 - e_4 + c_1 e_4 + e_1 e_4.\end{multline}But$$\det\begin{bmatrix}A&B\\C&D\end{bmatrix}=1 - c_1 - c_2 c_3 - c_4 + c_1 c_4$$and$$\det\begin{bmatrix}A&B\\E&0\end{bmatrix}=e_2 e_3 + e_1 e_4.$$

Answer 2

You can easily check facts about $2\times 2$ matrices and their determinants because $\det\begin{bmatrix} a & b \\ c & d \end{bmatrix}=ad-bc$, but you can't extend such facts to partitioned matrices, because $$\det\begin{bmatrix} A & B \\ C & D \end{bmatrix}=\det A \det(D-CA^{-1}B)$$if $A$ is nonsingular, or $$\det\begin{bmatrix} A & B \\ C & D \end{bmatrix}=\det D \det(A-BD^{-1}C)$$if $D$ is nonsingular (see e.g. Harville, Matrix Algebra From a Statistician's Perspective). In general, \begin{align*}\det\begin{bmatrix} A & B \\ C+E & D \end{bmatrix}&=\det A \det(D-(C+E)A^{-1}B)\\&\ne\det\begin{bmatrix}A & B \\ C & D\end{bmatrix}+\det\begin{bmatrix}A & B \\ E & 0\end{bmatrix}\\ &=\det A \det(D-CA^{-1}B)+\det A \det(0-EA^{-1}B) \end{align*}