I've been trying to find material on matrix calculus but it seems hard to find ones with understandable proofs.
I'm doing research work and I am trying to verify some computation. Suppose that I have a matrix $A= \left( \begin{array}{ccc}
\beta_{11} + c\beta_{12} +\beta_{13} & -c\beta_{12} & -\beta_{13} \\
-c\beta_{12} & c\beta_{12}+\beta_{22}+\beta_{23} & -\beta_{23} \\
-\beta_{13} & \beta_{23} & -\beta_{13}+\beta_{23}+\beta_{33} \end{array} \right) $ where $c$ is a constant, how do I evaluate $\frac{d}{d\beta_{12}}\det{(A)}$?
From what I have searched, if $A= \left( \begin{array}{ccc}
\beta_{11} & \beta_{12} & \beta_{13} \\
\beta_{21} & \beta_{22} & \beta_{23} \\
\beta_{31} & \beta_{32} & \beta_{33} \end{array} \right) $ , i.e. no 2 elements are identical, then $\frac{d}{d\beta_{12}}\det{(A)}=\det{(A)}\cdot A^{-1}_{12}$.
What about the former case? Is there some sort of product rule like in 1-variable calculus?
I forgot to mention, the above is just a simplified case of the problem I'm working on. For my case, the matrix $A$ has dimension 300×300
Thanks!
Best Answer
You can use the fact that $$\frac{\partial \det A}{\partial \alpha} = \det A \,\mathop{\rm tr} \left(A^{-1} \frac{\partial A}{\partial \alpha}\right), $$if $A$ is invertible.
Or more generally, you can use Jacobi's formula $$\frac{\partial \det A}{\partial \alpha}= \mathop{\rm tr} \left(\mathop{\rm adj}(A) \frac{\partial A}{\partial \alpha}\right).$$