Determinant of a square matrix with main diagonal of zeros?

Question

How can I show that the determinant of a square matrix A of dimension NxN with all elements equal to $-\delta$ except the main diagonal composed by zeros, is equal to $-(N-1)\times \delta^N$?

Answer 1

Your matrix is $$ \mathbf{A} = \delta (\mathbf{I} - \mathbf{1}\mathbf{1}^T) $$ where $\mathbf{I}$ is the identity matrix and $\mathbf{1}$ denotes the all-ones vector.

$$ \begin{align} \det(\mathbf{A}) &= \delta^N \det(\mathbf{I} - \mathbf{1}\mathbf{1}^T) \\ &= \delta^N \det(1 - \mathbf{1}^T\mathbf{1}) \\ &= \delta^N (1 - \mathbf{1}^T\mathbf{1}) \\ &= \delta^N (1 - N) \\ \end{align} $$

The first equality holds because $\det(\alpha\mathbf{A}) = \alpha^N \det(\mathbf{A})$ for any scalar $\alpha$ and $N$-by-$N$-matrix $\mathbf{A}$.

The second equality is due to Sylvester's determinant identity:

http://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem