I've tried the determinant of matrices with the form $$\left( \begin{matrix} x&amp;-1&amp;\cdots&amp;-1\\ -1&amp;x&amp;\cdots&amp;-1\\ \vdots&amp;\vdots&amp;\ddots&amp;-1\\ -1&amp;-1&amp;\cdots&amp;x \end{matrix} \right)_{n\times n}$$ i.e. with $x$ in the diagonal and $-1$ everywhere else. I've always get $$(x+1)^n-n(x+1)^{n-1}=(x+1)^{n-1}(x-n+1)$$ for various $n$ and I think it is a correct formula.  I think by induction will be a way but I don't know how. Any variant or counterexample are welcome.
I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a &amp; b &amp; \ldots &amp; b\\ b &amp; a &amp; \ldots &amp; b\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ b &amp; b &amp; \ldots &amp; a\end{bmatrix}$$  where $0 &lt; b &lt; a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated.
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