$$\mathbf{Xw}-\lambda_w \mathbf{1}=0$$
Where $\lambda_w$ is a scalar, $\mathbf{w}$ is a vector, and $\mathbf{X}$ is a symmetric p.d. matrix. It is known that $\mathbf{1'w}=1$ (but I don't think that matters too much for the present question).
This equation can be rewritten:
$$\mathbf{X}=\lambda_w \mathbf{D^{-1}}$$
where $\mathbf{D}=diag(\mathbf{w})$.
One more thing:
$$\mathbf{X}=\mathbf{\Sigma-\lambda_{\mu}R}$$
Where $\mathbf{\Sigma}$ and $\mathbf{R}$ are symmetric, p.d. matrices and $\lambda_{\mu}$ is another scalar.
The basic goal here is to solve for the $\mathbf{w}$ that make the first displayed equation true. $\lambda_w$ is known but $\lambda_{\mu}$ is not.
So, for example, in the simple case of a 2x2 matrix:
$$\begin{bmatrix} \Sigma_{11}-\lambda_{\mu}R_{11} & \Sigma_{12}-\lambda_{\mu}R_{12}\\ \Sigma_{21}-\lambda_{\mu}R_{21} & \Sigma_{22}-\lambda_{\mu}R_{22} \end{bmatrix}=\begin{bmatrix} \lambda_w/w_1 & 0\\ 0 & \lambda_w/w_2 \end{bmatrix}$$
Here you can see the off diagonal elements on the LHS have to sum to $0$, so:
$$\lambda_{\mu}=\frac{\Sigma_{ij}}{R_{ij}}$$
From the diagonal elements, on the other hand, we can get the following equation:
$$\lambda_{\mu}=\frac{1}{R_{ii}}\left(\Sigma_{ii}-\frac{\lambda_w}{w_i}\right)$$
Setting these two equations equal gives:
$$\frac{\lambda_w}{w_i}=\frac{\Sigma_{ii}R_{ij}-R_{ii}\Sigma_{ij}}{R_{ij}} \:\:\:\: \forall \: j \neq i$$
But the LHS are just the diagonal elements of the matrix $\lambda_w\mathbf{D^{-1}}$. Therefore the second displayed equation can be rewritten:
$$\mathbf{X}=\mathbf{\lambda_w D^{-1}}=\mathbf{Q}$$
where the elements of $\mathbf{Q}$ are defined:
$$Q_{ij}\left\{\begin{matrix} \frac{\Sigma_{ii}R_{ij}-R_{ii}\Sigma_{ij}}{R_{ij}} \:\:\:\: (\forall \: j \neq i) \:\:for \:\:on \:\: diagonal\:\: elements\\ \frac{\Sigma_{ii}R_{ij}-R_{ii}\Sigma_{ij}}{R_{ij}} \:\:\:\: (j = i) \:\:for \:\:off \:\:diagonal\:\:elements \end{matrix}\right.$$
For example in the 2x2 case,
$$\mathbf{Q}=\begin{bmatrix} \frac{\Sigma_{11}R_{12}-R_{11}\Sigma_{12}}{R_{12}} & \frac{\Sigma_{11}R_{11}-R_{11}\Sigma_{11}}{R_{12}} \\ \frac{\Sigma_{22}R_{22}-R_{22}\Sigma_{22}}{R_{22}} & \frac{\Sigma_{22}R_{21}-R_{22}\Sigma_{21}}{R_{21}} \end{bmatrix}$$
So the question is: can I derive the matrix $\mathbf{Q}$ using only matrix operations? Is there a more tidy way to define the elements of $\mathbf{Q}$?