TITLE: Determinant of a matrix with $t$ in all off-diagonal entries.
QUESTION [6 upvotes]: It seems from playing around with small values of $n$ that
$$
\det \left( \begin{array}{ccccc}
-1 & t & t & \dots & t\\
t & -1 & t & \dots & t\\
t & t & -1 & \dots & t\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
t  & t & t & \dots& -1
\end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)
$$
where $n$ is the size of the matrix.
How would one approach deriving (or at least proving) this formally?

Motivation
This came up when someone asked what is the general solution to:
$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$
and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.
So I wanted to ensure that these are also the only solutions for the case with more variables.

REPLY [3 votes]: Note that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\chi_{-T}$ of $-T$.
Since $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of $-T$ is $\chi_{-T}=X^{n-1}(X+nt)$ (the final factor must be that because the coefficient of $X^{n-1}$ in $\chi_{-T}$ is $\def\tr{\operatorname{tr}}-\tr(-T)=nt$). Now setting $X=-(1+t)$ gives 
$$
  \det(-(1+t)I-T)=(-1-t)^{n-1}(-1-t+nt)=(-1-t)^{n-1}(-1+(n-1)t)
$$
as desired.
This kind of question is recurrent on this site; see for instance Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$) and How to calculate the following determinants (all ones, minus $I$).