Recent content by canag

This theorem is true indeed, and doesn't even need A to be symmetric. Using : [tex] \frac{\partial}{\partial x} ln det A = \sum_{i,j} \frac{\partial a_{ij}}{x} \frac{\partial}{\partial a_{i,j}}[\tex] with : [tex]\frac{\partial }{\partial c_{ij}} ln det A = (A^{-1})_{ji}[\tex] you get ...