I have a matrix of:
$$
A = \begin{bmatrix}
a & b & b & \cdots & b \cr
-1 & 1 & 0 & \ddots & \vdots \cr
0 & -1 & 1 & \ddots & \vdots \cr
\vdots & \vdots & \ddots & \ddots & 0 \cr
0 & 0 & \cdots & -1 & 1 \cr
\end{bmatrix}_{N \times N}
$$
To calculate the determinant I followed the most general approach:
$$
\begin{aligned}
\Delta_A &= a \cdot \text{det}
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0 \cr
-1 & 1 & 0 & \cdots & \vdots \cr
0 & -1 & 1 & \cdots & \vdots \cr
0 & 0 & -1 & 1 & \vdots \cr
\vdots & \ddots & \ddots &\ddots & 1 \cr
\end{bmatrix}
– b \cdot
\text{det}
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 1 & 0 & 0 & 0 \\
0 & 0 & -1 & 1 & 0 & 0 \\
0 & 0 & 0 & -1 & 1 & \vdots \\
\vdots & 0 & 0 & 0 & \ddots & 1 \\
0 & 0 & 0 & 0 & 0 & -1
\end{bmatrix}
+ b \cdot
\text{det}
(A_{13}) + \cdots
\end{aligned}
$$
So the good news is the second matrix that I faced is a diagonal matrix consisting of either 1 or -1 on the diagonal. Actually, it's not a diagonal matrix but it's a diagonal matrix with non-zero off-diagonal values. I numerically calculate the determinant of such a matrix (up to size 100) and it seems that the determinant is no different from the determinant of a simple diagonal matrix:
$\text{det} A = \prod_{i=1}^{N} A_{ii}$
I wonder if there's any proof for that.
Best Answer
Seems to me that the matrix is a particular upper Hessenberg so the closed form of its determinant in this case should be $$\sum_{k=0}^{n-1} \left(a_k\prod_{i=1}^{k} y_i\prod_{i=k+1}^{n-1} x_i\right)$$
where $a_k = \begin{cases}a & k = 0 \\ b & k \geq 1\end{cases}$ and $x_i = y_i = 1$ for each $i$. It follows that the formulas simplifies to $$\sum_{k=0}^{n-1} a_k = a + \sum_{k=1}^{n-1} a_k = a + \sum_{k=1}^{n-1} b = a+(n-1)b$$