Determinant of a special kind of block matrix I have a $2\times2$ block matrix $M$ defined as follows:
$$\begin{pmatrix}X+|X| & X-|X| \\  Y-|Y| & Y+|Y|\end{pmatrix}$$
where $X$ and $Y$ are $n\times n$ matrices and $|X|$ denotes the modulus of the entire matrix $X$ that essentially comprises modulus of individual elements of $X$.
How may I find the determinant of the matrix $M$ in terms of $X$ and $Y$? Looking for a simplified solution?

I shall assume that $X+|X|$ is invertible, although a similar solution exists under the assumption that $Y+|Y|$ is. I shall use $A,B,C,D$ to denote the respective block matrices in your problem to avoid giant equations. The decomposition $$M = \begin{pmatrix}A & B \\  C & D\end{pmatrix} = \begin{pmatrix}A & 0\\  C & I\end{pmatrix} \begin{pmatrix}I & A^{-1}B\\  0 & D-CA^{-1}B\end{pmatrix} = ST$$ can be verified by simple matrix multiplication (noting that matrix multiplication for block matrices works like multiplying matrices over any other noncommutative ring). The key fact is that $\det(S) = \det(A)$ and $\det(T) = \det(D-CA^{-1}B)$. I shall only prove the first equality (or rather a stronger statement where $I$ is not necessarily the same size as $A$ but the block matrix is still square), as the second can be proved similarly. If $A$ is $1\times 1$, the equation follows from the fact that $S$ is triangular and the product along its diagonal is $\det(A)$. If we assume that it holds for any $n\times n$ matrices $A,C$ then we can apply the Laplace formula to get $$\det(S) = \sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(N_{1j})$$ where $N_{1j}$ is the matrix that results from deleting the first row and $j^{th}$ column of $S$. These matrices satisfy the inductive hypothesis (the identity matrix has not been touched), and so $\det(N_{1j}) = \det(M_{1,j})$ where $M_{1j}$ is the matrix that results from deleting the first row and $j^{th}$ column of $A$. Using the Laplace formula again gives $$\sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(N_{1j}) = \sum\limits_{j=1}^{n+1} (-1)^{j+1}a_{1j}\det(M_{1j}) = \det(A)$$ completing the proof. Since $\det(M) = \det(S)\det(T)$, this gives us $$\det(M) = \det(A)\det(D-CA^{-1}B)$$
