How can $\frac{\partial A^{-1}}{\partial A}$ ever be $A^{-1}\mathbb{I}A^{-1}$? $A$ being a symmetric second-order tensor. On several spots including this wikipedia article I've seen
$$\left(\frac{\partial A^{-1}}{\partial A}\right)_{ijkl}=
{\frac {\partial A_{ij}^{-1}}{\partial A_{kl}}}=
-{\frac {1}{2}}\left(A_{ik}^{-1}~A_{jl}^{-1}+A_{il}^{-1}~A_{jk}^{-1}\right).$$
On several other spots including this paper (picture below) I've seen the formula
$$\frac{\partial A^{-1}}{\partial A} =  -A^{-1}\mathbb{I}A^{-1},$$
About the notation of the paper, I know for sure that $\partial_{\mathbf{b}}(\mathbf{b}^{-1})=\frac{\partial \mathbf{b}^{-1}}{\partial \mathbf{b}}$, I know for sure that $\mathbf{b}$ is a symmetric second-order tensor and I'm pretty sure $\mathbb{I}$ is supposed to be the fourth order symmetric unit tensor $\frac{1}{2}\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)(\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l)$, although I couldn't find an explicit definition anywhere.
Trying to derive
${\frac{1}{2}}\left(A_{ik}^{-1}~A_{jl}^{-1}+A_{il}^{-1}~A_{jk}^{-1}\right)$ starting from $A^{-1}\mathbb{I}A^{-1}$, I get
$$\frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})
A_{ab}^{-1}(\textbf{e}_a\otimes\textbf{e}_b)
(\mathbf{e}_i\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_l)
(\textbf{e}_c\otimes\textbf{e}_d)A_{cd}^{-1}=
\\  
\frac{1}{2}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})
A_{ab}^{-1}\delta_{bi}
(\mathbf{e}_a\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_d)
\delta_{cl}A_{cd}^{-1} =
\\
\frac{1}{2}(\delta_{ik}\delta_{jl}\delta_{bi}\delta_{cl}+\delta_{il}\delta_{jk}\delta_{bi}\delta_{cl})
A_{ab}^{-1}
(\mathbf{e}_a\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_d)
A_{cd}^{-1} = 
\\ \frac{1}{2}(A_{ak}^{-1}A_{jd}^{-1}+A_{ai}^{-1}A_{id}^{-1}\delta_{jk})(\mathbf{e}_a\otimes\mathbf{e}_j\otimes\mathbf{e}_k\otimes\mathbf{e}_d).
$$
And that's where I'm stuck. The first term $A_{ak}^{-1}A_{jd}^{-1}$ checks out perfectly, but the second one is $A_{ai}^{-1}A_{id}^{-1}\delta_{jk}$ even though it should be $A_{ad}^{-1}~A_{jk}^{-1}$, and I have no idea what to do with it.
Can I simplify this further to get the desired result? Is there a mistake in the paper? Or did I make a mistake anywhere or get some notation detail wrong?

 A: There's nothing really crucial about matrices (which are used to represent rank 2 tensors in coordinates) here, and you should think of this formula as a non-commutative version of $(1/x)' = -1/x^2$, where each inverse goes to one side.
Let $G$ be a Lie group and $\iota\colon G \to G$ be the inversion map $\iota(g) = g^{-1}$. Let's show that ${\rm d}\iota_g(v) = -g^{-1}vg^{-1}$, where we use the usual abuses of notation $av$ for ${\rm d}(L_a)_g(v)$ and $vb = {\rm d}(R_b)_g(v)$, where $v\in T_gG$.
If $m\colon G\times G \to G$ is the group multiplication, $m(g,h) = gh$, then we have that ${\rm d}m_{(g,h)}(v,w) = gw+hv$. We use this to implicitly differentiate the relation $m(g,\iota(g)) = e$, relative to $g$. This gives: $$vg^{-1} + g{\rm d}\iota_g(v) = 0 \implies {\rm d}\iota_g(v) = -g^{-1}vg^{-1},$$as required. When $G$ is a subgroup of ${\rm GL}(n,\Bbb R)$, the left and right multiplications are restrictions of linear maps, so the abuse of notation mentioned above is no longer an abuse of notation, but the actual truth. Namely, if $\iota(A) = A^{-1}$, then we do get $${\rm d}\iota_A(H) = -A^{-1}HA^{-1}$$Since the $H$ in the middle is the identity map evaluated at $H$, the notation $$\frac{\partial A^{-1}}{\partial A} = -A^{-1} \Bbb I A^{-1}$$is somewhat justified (although I'm not a fan of it).
