Let $C_\omega = (I-\omega D^{-1}L)^{-1}((1-\omega)I+\omega D^{-1}R)$ then $\det(C_\omega) = (1-\omega)^n$
(Where $C_\omega\in \mathbb{R}^{n\times n}$, $R$ is upper triangular, $L$ is lower triangular and $D$ is a diagonal matrix)
Could someone explain why this is the case?
I know:
$$\begin{align} \det(C_\omega) &= \det(I-\omega D^{-1}L)^{-1} \cdot \det ((1-\omega)I+\omega D^{-1}R)\\ & = [\det(I-\omega D^{-1}L)]^{-1}\cdot \det((1-\omega)I+\omega D^{-1}R)\\ & = \ldots? \end{align}$$
Okay, after staring at it some time, I managed to see why.
Since $L$ is a lower triangular matrix, $R$ is an upper triangular matrix. Then $$\omega D^{-1}L\text{ is lower triangular}$$ So $I-\omega D^{-1}L\text{ is lower triangular}$ is lower triangular with 1's on the main diagonal which means it's determinant is 1.
The same reasoning holds for the second part, but now $1-\omega$ resides on the diagonal, which results in a determinant of $(1-\omega)^n$.