I am working through some notes on representation theory, and in the questions I have come across this question:
A group of order $168$ has $6$ conjugacy classes of sizes $1, 21, 42, 56, 24, 24$. Three representations of this group are known, with their character given as:
$A: 14, 2, 0, -1, 0, 0 $
$B: 15, -1, -1, 0, 1, 1 $
$C: 16, 0, 0, -2, 2, 2 $
(So on the class of size 1, $Tr(A(x)) = 14$, on the class of size 21, $Tr(A(x)) = 2$ and so on)
There is also a hint that you can assume that $\sqrt{7}$ is not in $\Bbb{Q}(\zeta)$ where $\zeta$ is a seventh root of unity
I first computed the inner products of each representation with itself and with each other:
$\langle A, B\rangle = 1$, $\langle A, C\rangle = 2$, $\langle C, B\rangle = 2$
$\langle A, A\rangle = 2$, $\langle B, B\rangle = 2$, $\langle C, C\rangle = 4$
So $A, B$ are each the sum of two irreducible characters, and $C$ is either twice an irreducible character or the sum of $4$ irreducible characters, but if it were the sum of $4$ irreducible characters, $C - B$ would be a degree $1$ character, and therefore irreducible, but also the sum of $2$ irreducible characters, contradiction, so $C = 2\chi_1$, and we have our first irreducible character.
The inner products with $A, B$ show that $\chi_1$ is one of the two characters that sum with them, so $\chi_2 = A - \chi_1$ and $\chi_3 = B - \chi_1$ are irreducible characters, so including the trivial character, we have 4 out of 6 characters:
$\chi_0: 1, 1, 1, 1, 1, 1$
$\chi_1: 8, 0, 0, -1, 1, 1$
$\chi_2: 6, 2, 0, 0, -1, -1$
$\chi_3: 7, -1, -1, 1, 0, 0$
Now as $168 = \sum_0^5 \chi_i(1)^2$, we have that the remaining representations are both degree $3$.
I know that we can get these by column and row orthogonality and solving the ridiculous amount of simultaneous equations that we get, and using products of characters I have that:
$\chi_4 + \chi_5: 6, -2, 2, 0, -1, -1$
But is there a quicker way to get them?