I would like to follow up on one of my previous questions, where I have the following block matrix with dimensions
$ [M] := \begin{bmatrix} A & B\\ C & D \end{bmatrix} := \begin{bmatrix} (1\times1) & (1\times n)\\ (n\times1) & (n\times n) \end{bmatrix}$
To calculate the determinant, I can use the following formula (see previous question for clarification)
$Det[M] = a^{1-n}Det[aD-CB]$ where "a" is a scalar
My question is, if I have an operator acting on the determinant,
$\frac{\partial}{\partial t}Det[M] = \frac{\partial}{\partial t}(a^{1-n}Det[aD-CB])$
do the following properties hold when applied with the formula...
- product rule
$\frac{\partial}{\partial t}Det[M] = a^{1-n}\frac{\partial}{\partial t}(Det[aD-CB])+Det[aD-CB]\frac{\partial}{\partial t}(a^{1-n})$
- if so, in the first term on the right hand side, can I distribute the operator inside the expression
$\frac{\partial}{\partial t}(Det[aD-CB])=Det[\frac{\partial}{\partial t}(aD)-\frac{\partial}{\partial t}(CB)]$
I'm just not sure how to go about solving
$\frac{\partial}{\partial t}Det[M] = \frac{\partial}{\partial t}(a^{1-n}Det[aD-CB])$
The notation $AD$ is a very bad one ($AD$ is not a matrix product). Let $A=a$.
It remains to calculate $(\det(aD-CB))'=\det(aD-CB)trace((aD-CB)'(aD-CB)^{-1})$
$=\det(aD-CB)trace((a'D+aD'-C'B-CB')(aD-CB)^{-1})$.