I want to solve the $\Lambda$ from the following equation. $\frac{\partial \det(\Lambda)^{C_{1}} e^{C_{2} tr(w^{-1}\Lambda+\Lambda S)}}{\partial \Lambda} = 0$ where $C_{1}, C_{2}$ is some constant and $\Lambda$, $w^{-1}$, $S$ are all $2x2$ matrix.
I have read the matrix calculus on wiki.
I acknowledge it is a scalar by matrix. And I want to use the denominator layout.
I write it to $\frac{\partial{\det(\Lambda)^{C_{1}}}}{\partial{\Lambda}} e^{C_{2}tr(w^{-1}\Lambda+\Lambda S)} + \det(\Lambda)^{C_{1}}\frac{\partial{e^{C_{2} tr(w^{-1}\Lambda+\Lambda S)}}}{\partial{\Lambda}}$
I want to calculus the two partial derivate respectively. But I cannot find the identity matrix about $\frac{\det(\Lambda)^{C_{1}}}{\partial{\Lambda}}$, I only find the $\frac{\det(\Lambda^{C_{1}})}{\partial{\Lambda}}$.
I cannot find any identity matrix about $\frac{\partial{e^{C_{2} tr(w^{-1}\Lambda+\Lambda S)}}}{\partial{\Lambda}}$.
Please give me some resources for the identity matrix or how to solve it.(I know the calculus but is unfamiliar to matrix calculus. I had solved it when I written it to one dimension equation but it is too demanding)
Thank you.
Let $$\eqalign{ A &= \Lambda \cr b &= C_1 \cr c &= C_2 \cr e &= \exp\big(C_2\,{\rm tr}(w^{-1}A+AS)\big) \cr t &= \det(A) \cr }$$ Then write the objective function and find its differential $$\eqalign{ f &= et^b \cr df &= ebt^{b-1}\,dt + t^b\,de \cr &= ebt^b\,(\frac{dt}{t}) + t^b\,ce\,d\,{\rm tr}(w^{-1}A+AS) \cr &= ebt^b\,d\log(t)) + t^b\,ce\,d{\rm tr}(w^{-1}A+AS) \cr }$$ If $A$ is non-singular then Jacobi tells us that $$\log(\det(A))={\rm tr}(\log(A))$$ Substituting this result, we get $$\eqalign{ \Big(\frac{1}{et^b}\Big)df &= b\,d{\rm tr}(\log(A)) + c\,d{\rm tr}(w^{-1}A+AS) \cr &= b\,A^{-T}:dA + c\,(w^{-1}+S):dA \cr\cr \frac{\partial f}{\partial A} &= et^b\Big(b\,A^{-T} + c\,(w^{-1}+S)\Big) \cr }$$ Setting the gradient to zero and solving for $A$ yields $$\eqalign{ b\,A^{-T} &= -c\,(w^{-1}+S) \cr A &= -\frac{b}{c}\,(w^{-1}+S)^{-T} \cr }$$