My question comes from a section of Representations and Characters of Groups, by James & Liebeck. They discuss the following example in the chapter $15$ on The Number of Irreducible Characters
We shall see in Section $18.4$ that there is a certain group $G$ of order $12$ which has exactly six conjugacy classes with representatives $g_1, \ldots, g_6$ (where $g_1=1$), and six irreducible characters $\chi_1, \ldots, \chi_6$ given as follows:
Suppose we are given characters $\chi$ and $\psi$ of $G$ as follows:
Then it is easy to spot that $$\chi = \chi_2+\chi_6, \quad \psi=\chi_1+\chi_2+\chi_3+\chi_4.$$ For example, the second entry in the row vector for $\chi$ is equal to minus the first entry. Inspecting the values of the irreducible characters $\chi_i$, we see that $\chi$ must be a combination of $\chi_2, \chi_4$ and $\chi_6$. The correct answer now comes quickly to mind.
I don't see what's so "easy to spot" and how the answer "comes quickly to mind." I understand the answer that they got, but I must be missing some trick that makes this example so "easy."