$$M = \begin{bmatrix} X & \begin{bmatrix} I\\I\\-I\\-I\end{bmatrix}\\ \begin{bmatrix}-I&-I&I&I\end{bmatrix}&O\end{bmatrix}$$
Is there an easy way to compute the determinant of the block matrix above? Or, at least, can I know if $\det (M) > 0$ if $X$ is positive definite?
$X$ is, for example, an $8 \times 8$ matrix while $I$ is the $2 \times 2$ identity and $O$ is the the $2 \times 2$ zero matrix.
$$\mathrm M := \left[\begin{array}{c|c} \,\mathrm X & \mathrm C\\ \hline -\mathrm C^\top & \,\,\mathrm O_2\end{array}\right]$$
where $n \times n$ matrix $\rm X$ is positive definite and, thus, invertible. Using Gaussian elimination,
$$\left[\begin{array}{c|c} \mathrm I_n & \mathrm O_{n \times 2}\\ \hline \mathrm C^\top \mathrm X^{-1} & \mathrm I_2\end{array}\right] \left[\begin{array}{c|c} \,\mathrm X & \mathrm C\\ \hline -\mathrm C^\top & \,\,\mathrm O_2\end{array}\right] = \left[\begin{array}{c|c} \mathrm X & \mathrm C\\ \hline \quad\mathrm O_{2 \times n} & \,\,\mathrm C^\top \mathrm X^{-1} \mathrm C\end{array}\right]$$
Since the determinant of a block triangular matrix is the product of the determinants of the diagonal blocks,
$$\det (\mathrm M) = \det (\mathrm X) \cdot \det \Big(\mathrm C^\top \mathrm X^{-1} \mathrm C \Big)$$
where the Schur complement $\mathrm C^\top \mathrm X^{-1} \mathrm C$ is a $2 \times 2$ matrix (whose determinant is easy to compute). Note that $\det (\mathrm X) > 0$ and that $n \times 2$ matrix $\mathrm X^{-1} \mathrm C$ can be computed using Gaussian elimination.
Using SymPy to verify:
Building the elimination matrix,
Performing Gaussian elimination,