I have some problems with finding the determinant of the following matrix.
I have tried a simple Laplace expansion on the first row, but I have a feeling I am missing some simple trick here. $$ \begin{bmatrix} y_{n} & 0 & 0 & \dots &0 & y_{1} \\ 0& y_{n} & 0 & \dots &0 & y_{2} \\ 0&0& y_{n} & \dots &0 & y_{3} \\ \vdots & \vdots & \vdots & \ddots & \vdots &\vdots \\ 0 & 0 & 0&\dots & y_{n} & y_{n-1} \\ -y_{n} & -y_{n} & -y_{n} & \dots &-y_{n} & 1- \sum_{i=1}^{n-1}y_i \end{bmatrix} $$
Besides that, while I was doing this I got interested in computing the determinant of arrowhead matrices and I can not find something that helps.
Starting with dim. 2 and 3, it is not difficult to deduce, and then demonstrate that it holds for all 2 <= dim., that the LU decomposition of your matrix is
$$ \left[ {\begin{array}{*{20}c} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ { - 1} & { - 1} & \cdots & { - 1} & 1 \\ \end{array} } \right]\;\left[ {\begin{array}{*{20}c} {y_n } & 0 & \cdots & 0 & {y_1 } \\ 0 & {y_n } & \cdots & 0 & {y_2 } \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & {y_n } & {y_{n - 1} } \\ 0 & 0 & \cdots & 0 & 1 \\ \end{array} } \right] $$ By the way, many times "arrow-head" m. decompose LU in "half-arrow-head".