I am working with a system of linear equations $Ax=b$ where $A$ is given by the character table of a finite group $G$ (actually, the group of permutations $\Sigma_k$). In other words, the equations are $$ \sum_{R\text{ irred}} \chi_R(\sigma)x_R = b_\sigma,$$ where $R$ runs in the set of irreducible representations of $G$ and $b_\sigma=b_{\sigma'}$ if $\sigma$ and $\sigma'$ are conjugated.
I am mostly interested in the case $G=\Sigma_k$. In this case it is easy to find the variable corresponding to the trivial and alternating/sign representations, but the remaining variables are tricky.
To use Cramer's method I would need to know the minors of the character table, but I do not know any reference for that and the computation seems tedious and obscure in general. I hope there is a clean way of doing it, maybe with all the elements of the group instead of the conjugacy classes.
On the other hand, the determinant of $A$ would be easier, for example it is given in Determinant of the character table of a finite group $G$
The trivial variable is equal to $$ 1/k!\sum_{\sigma}b_\sigma $$ and the sign variable is equal to $$ 1/k!\sum_{\sigma}sgn(\sigma)b_\sigma $$
It was way simpler than I though, so I put the answer here myself:
As the characters of irreducible representations are pair-wise orthogonal, one can simply take the sum of $$ \overline{\chi_{R_0}(\sigma)}\sum_{R \text{ irred}}\chi_R(\sigma)x_R= \overline{\chi_{R_0}(\sigma)}b_\sigma,$$ for some irreducible representation $R_0$. It yields $$ |G|x_{R_{0}} = \sum_\sigma \overline{\chi_{R_0}(\sigma)}b_\sigma.$$