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 $1$.
 A: $$
 \begin{vmatrix}
  1 & -1 & -1 & -1 & \cdots & -1 \\
  1 &  1 &  0 &  0 & \cdots &  0 \\
  1 &  0 &  1 &  0 & \cdots &  0 \\
  1 &  0 &  0 &  1 &  &  0 \\
  \vdots & \vdots &  \vdots &   & \ddots &  \vdots \\
  1 & 0 & 0 & \cdots & 0 & 1
 \end{vmatrix} = 
 \begin{vmatrix}
  n & 0 & 0 & 0 & \cdots & 0 \\
  1 &  1 &  0 &  0 & \cdots &  0 \\
  1 &  0 &  1 &  0 & \cdots &  0 \\
  1 &  0 &  0 &  1 &  &  0 \\
  \vdots & \vdots &  \vdots &   & \ddots &  \vdots \\
  1 & 0 & 0 & \cdots & 0 & 1
 \end{vmatrix} =
n
$$
Note that in a triangular matrix its determinant equals the product of the diagonal entries.
A: Yes: Use Gauss-Jordan row elimination starting from the lower right and working your way up with pivots always on the diagonal. This elimination will not affect the determinant unless you have to divide  a pivot row by something other than 1 (in which case you have multiplied the determinant by that same factor.
After the first step, which only has to add the last row to the first, the matrix is in the same except there is a 2 in the upper left, and a zero in the upper right. 
After the next step, there is a 3 in the upper left, and a zero in the upper right and in
$A_{n,n-1}$. And so forth. 
You end up with $n$ in the upper left.
As a final step, row eliminate using the first row as a pivot to wipe out the rest of the first column without changing the determinant.
A: 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.
