I am looking to show that the Ricci tensor of the first kind, $R_{i j}$ obtained by retracting the Riemann tensor of the first kind, via $R_{i j} = R^k_{i j k}$ is symmetry. To do this, I have shown that
$R_{i j} = \frac{\partial^2}{\partial x^i \partial x^j} \log \left( \sqrt{|g|} \right) - \frac{1}{\sqrt{ |g| } } \frac{\partial}{\partial x^r} \left( \sqrt{ |g| } \Gamma^r_{i j} \right) + \Gamma^r_{i s} \Gamma^s_{r j},$
where $g = \det G = \det g_{i j}$, as symmetry easily follows from this,
but in doing so I have used the fact that $\frac{\partial}{\partial x^i} \left( \log \det A \right) = b_{s r} \frac{\partial a_{r s}}{\partial x^i}$, where $A = \left[ a_{i j} (x) \right]$ and $B = \left[ b_{i j} (x) \right]$, which I have so far been unable to prove.
I have been given the following hint to do so, however:
"Start by justifying the following chain of identities: $\frac{\partial}{\partial x^i} \left( \log \det A \right) = \frac{1}{\det A} \frac{\partial}{\partial x^i} \left( \det A \right) = \frac{1}{\det A} \frac{\partial}{\partial a_{r s}} \left( \det A \right) \frac{\partial a_{r s}}{\partial x^i} = \frac{ \left( \textrm{cof } A \right)_{ r s}}{\det A} \frac{\partial a_{r s}}{\partial x^i} = b_{s r} \frac{\partial a_{r s}}{\partial x^i}$, where $A$ and $B$ are as defined previously."
I think the first identity in the chain can be justified by the usual rules of differentiating a $\log$, the second via the 'chain rule' but I am unsure of the last two. I'd very much appreciate a hint. :)