Suppose we have a network of circuit nodes connected to each other via two-terminal mutual capacitances $C_{m,ij}$. Is there a recipe for finding the capacitance matrix for this network, and for eliminating certain nodes that are allowed to float?
This question was inspired by Total capacitance from a capacitance matrix.
The capacitance matrix is defined by $$\begin{pmatrix}Q_1 \\ Q_2 \\ \vdots \\ Q_n \end{pmatrix} = \begin{pmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & \cdots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{pmatrix} \, , $$ i.e. given a set of voltages imposed on each conductor, the capacitance matrix tells us the resulting charge on each conductor. But we can also write the charges in terms of the mutual capacitances. For example, \begin{align} Q_1 &= C_{m,11}V_1 + C_{m,12}(V_1 - V_2) + C_{m,13}(V_1 - V_3) + \cdots \\ &= (C_{m,11} + C_{m,12} + C_{m,13} + \cdots )V_1 - C_{m,12}V_2 - C_{m,13}V_3 - \cdots \, . \end{align} Therefore, the capacitance matrix is $$ \begin{pmatrix} \sum_{i=1}^n C_{m,1i} & -C_{m,12} & \cdots & -C_{m,1n} \\ -C_{m,21} & \sum_{i=1}^n C_{m,2i} & \cdots & -C_{m,2n} \\ \vdots & \vdots & \ddots & \vdots \\ -C_{m,n1} & -C_{m,n2} & \cdots & \sum_{i=1}^n C_{m,ni} \end{pmatrix} \, . $$
There is an algorithm to reduce a capacitance matrix, i.e. to find the capacitance matrix in the case that we choose one conductor to float (we do not impose any particular voltage on that node). Suppose we have the matrix $$ C = \begin{pmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & & C_{2n} \\ \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{pmatrix} $$ and suppose we wish to know the effective capacitance matrix $\tilde{C}_{\text{drop }k}$ representing all nodes except for node $k$, where node $k$ is left floating. The way to do this is first draw this diagram $$ \begin{pmatrix} C_{11} & \cdots & \color{red}{C_{1k}} & \cdots & C_{1n} \\ \vdots & & \color{red}{\vdots} & & \vdots \\ \color{red}{C_{k1}} & \color{red}{\cdots} & \color{red}{C_{kk}} & \color{red}{\cdots} & \color{red}{C_{kn}} \\ \vdots & & \color{red}{\vdots} & & \vdots \\ C_{n1} & \cdots & \color{red}{C_{nk}} & \cdots & C_{nn} \end{pmatrix} $$ We're going to delete $k^\text{th}$ row and column, i.e. the stuff shown in red. We're also going to adjust the remaining matrix elements. Consider for example the element $C_{11}$. We draw a horizontal and vertical line, starting at that element, until we hit red, giving us two red values $\color{red}{C_{1k}}$ and $\color{red}{C_{k1}}$. We adjust $C_{11}$ by subtracting the product of the two red values divided by $C_{kk}$, like this $$ C_{11} \rightarrow C_{11} - C_{1k}C_{k1} / C_{kk} \, .$$ Let's do another one. Consider $C_{n1}$. We draw horizontal and vertical lines hitting $\color{red}{C_{k1}}$ and $\color{red}{C_{nk}}$, and so we get $$C_{n1} \rightarrow C_{n1} - C_{k1}C_{nk} / C_{kk} \, .$$ Do this for all remaining elements and you will have constructed the desired reduced matrix with node $k$ floating.
Let's do a simple example. Consider two conductors 1 and 2 with mutual capacitance $C_{m,12}$, and mutual capacitances to ground $C_{m,11}$ and $C_{m,22}$. Let us find the total capacitance to ground for conductor 1.
First, using our rule to construct the capacitance matrix from the mutual capacitances, we have $$C = \begin{pmatrix} C_{m,11} + C_{m,12} & - C_{m,12} \\ - C_{m,21} & C_{m,21} + C_{m,22} \end{pmatrix} \, . $$ Then, using our rule to eliminate node 2, we find
\begin{align} C_{1, \text{to ground}} = \tilde{C}_{\text{drop }2} &= C_{11} - \frac{C_{21}C_{12}}{C_22} \\ &= C_{m,11} + C_{m,12} - \frac{(-C_{m,21})(-C_{m,12})}{C_{m,21} + C_{m,22}}\\ &= C_{m,11} + C_{m,12} - \frac{C_{m,12}^2}{C_{m,12} + C_{m,22}} \, . \end{align} To verify that this is correct, let's compute the capacitance of node 1 to ground using parallel/series rules. \begin{align} C_{1, \text{to ground}} &= C_{m,11} + C_{m,12} || C_{m,22} \\ &= C_{m,11} + \frac{C_{m,12}C_{m,22}}{C_{m,12} + C_{m,22}} \\ &= C_{m,11} + C_{m,12} - C_{m,12} + \frac{C_{m,12}C_{m,22}}{C_{m,12} + C_{m,22}} \\ &= C_{m,11} + C_{m,12} + \frac{C_{m,12}C_{m,22} - C_{m,12}(C_{m,12} + C_{m,22})}{C_{m,12} + C_{m,22}} \\ &= C_{m,11} + C_{m,12} - \frac{C_{m,12}^2}{C_{m,12} + C_{m,22}} \end{align} which agrees with the result we got using the matrix rules.