It's for a proof of Cayley's Formula, I know I'm being dumb and can't see it, how do I find the determinant of this $(n-1)\times (n-1)$ matrix where the diagonal entries are $n-1$ and the off diagonal all $-1$?
\begin{pmatrix}
n-1 & -1 & \cdots & -1 \\
-1 & n-1 & \cdots & \vdots \\
\vdots & \vdots & \ddots & \vdots \\
-1 & \cdots & \cdots & n-1
\end{pmatrix}
Best Answer
Replace the first column by the sum of all the columns: the wanted determinant is equal to $$\Delta_n:=\det\begin{pmatrix} 1 & -1 & \cdots & -1 \\ 1 & n-1 & \cdots & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \cdots & \cdots & n-1 \end{pmatrix}.$$ Now, for each $j\in\{2,\dots,n\}$, take the column $C_j$ and replace it by $C_j+C_1$ to obtain $$\Delta_n=\det\begin{pmatrix} 1 & 0 & \cdots & 0 \\ 1 & n & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & \cdots & n \end{pmatrix}.$$ The determinant of a triangular matrix is easier to compute.