Given a square matrix, all elements outside of the main diagonal being equal to $1,$ what is its determinant?
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See http://math.stackexchange.com/questions/219731/determinant-of-rank-one-update-of-identity-matrix (The chosen answer handles your case, even though the question doesn't) – deinst Apr 17 '15 at 17:33
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@deinst I think that covers it. Would you mind copying the relevant part of that answer into an answer to this question so I can mark this as accepted? I can do this myself, too, if you don't want to. – FUZxxl Apr 17 '15 at 17:37
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@deinst Strictly speaking, the case that at least one of the diagonal entries is $1$ (i..e, singular $A$) is not covered in (6.2.3) – Hagen von Eitzen Apr 17 '15 at 17:40
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@HagenvonEitzen When exactly one entry in the diagonal is $1$, the result is $\prod_i(a_{ii}-[a_{ii}\neq1]),$ where $[p]$ is the Iverson bracket. If more than one entry is $1$, the result is $0$. – FUZxxl Apr 17 '15 at 19:23
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Let the matrix be \begin{align} A & = \begin{bmatrix} d_1 & 1 & 1 & \cdots & 1\\ 1 & d_2 & 1 & \cdots & 1\\ 1 & 1 & d_3 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & d_n \end{bmatrix}\\ & = \begin{bmatrix} d_1-1 & 0 & 0 & \cdots & 0\\ 0 & d_2-1 & 1 & \cdots & 0\\ 0 & 0 & d_3-1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & d_n-1 \end{bmatrix}+ \begin{bmatrix} 1\\ 1\\ 1\\ \vdots\\ 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \end{bmatrix}\\ & = D+uu^T \end{align} where $D=\begin{bmatrix} d_1-1 & 0 & 0 & \cdots & 0\\ 0 & d_2-1 & 1 & \cdots & 0\\ 0 & 0 & d_3-1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & d_n-1 \end{bmatrix}$ and $u^T=\begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \end{bmatrix}$.
If $D$ is invertible, from [Sylvester determinant theorem][1], we have $$\det(D+uu^T) = \det(D)\det(I+D^{-1}uu^T) = \det(D) \det(I_{1 \times 1} + u^TD^{-1}u)$$
This gives us that the determinant is $$\prod_{k=1}^n (d_k-1)\left(1+\sum_{i=1}^n\dfrac1{d_i-1}\right) = \prod_{k=1}^n (d_k-1) + \sum_{i=1}^n \prod_{\overset{k=1}{k \neq i}}^n (d_k-1)$$ [1]: http://en.wikipedia.org/wiki/Sylvester's_determinant_theorem
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1The proof doesn't hold when D isn't invertible. Can this situation be proved separately? – Alireza Honarvar May 07 '21 at 17:39