<s>[INST] Calculation of determinant of an arrowhead matrix Is there any easier way to make sure the determinant of the following $n \times n$ matrix is $n$?
$$\begin{vmatrix}
  1 & -1 & -1 & -1 & \cdots & -1 \\
  1 &  1 &  0 &  0 & \cdots &  0 \\
  1 &  0 &  1 &  0 & \cdots &  0 \\
  1 &  0 &  0 &  1 & \cdots &  0 \\
  \vdots & \vdots &  \vdots & \vdots  & \ddots &  \vdots \\
  1 & 0 & 0 & 0 &\cdots  & 1
 \end{vmatrix} = n$$
I figured it with a smaller dimension and it indeed produces the determinant that is the size of dimension. I tried to do a cofactor expansion with the first row, and each term produces the determinant of $1$ and if you sum them up, then the total determinant will be $n$. But the sign change for each cofactor is confusing, and it is not easily seen that each cofactor term is actually positive $1$.
 [/INST] We know that for $n=2$ the statement is true. We'll prove the rest by induction.
We can compute the determinant based on the last row. The result is the sum of two determinants  
$\ \  \ \begin{vmatrix}
  1 & -1 & -1 & -1 & \cdots & -1 \\
  1 &  1 &  0 &  0 & \cdots &  0 \\
  1 &  0 &  1 &  0 & \cdots &  0 \\
  1 &  0 &  0 &  1 &  &  0 \\
  \vdots & \vdots &  \vdots &   & \ddots &  \vdots \\
  1 & 0 & 0 & \cdots & 0 & 1
 \end{vmatrix}=(-1)^{n+1}
\begin{vmatrix}
  -1 & -1 & -1 & -1 & \cdots & -1 \\
  1 &  0 &  0 &  0 & \cdots &  0 \\
  0 &  1 &  0 &  0 & \cdots &  0 \\
  0 &  0 &  1 &  0 &  &  0 \\
  \vdots & \vdots &  \vdots &   & \ddots &  \vdots \\
  0 & 0 & 0 & \cdots & 1 & 0
 \end{vmatrix}
+\begin{vmatrix}
  1 & -1 & -1 & -1 & \cdots & -1 \\
  1 &  1 &  0 &  0 & \cdots &  0 \\
  1 &  0 &  1 &  0 & \cdots &  0 \\
  1 &  0 &  0 &  1 &  &  0 \\
  \vdots & \vdots &  \vdots &   & \ddots &  \vdots \\
  1 & 0 & 0 & \cdots & 0 & 1
 \end{vmatrix},$
where the second  term is of the form of the original determinant. By hypothesis the value of this second determinant is $n-1$. 
From here it would be enough to show that the determinant of the first term is $-1$ for even $n$'s and $+1$ for odd $n$'s. For $n=2 \text {  and } 3$ this is true.
In the general case we can do the first term based on the last row. Surprisingly, by hiding the last but one column and the last row we get a determinant of the form of its predecessor. This completes the proof.
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