How to calculate determinant of matrix with 9's everywhere off-diagonal?

Question

I need help about this question for my algebra exam. I would be very thankful if someone could help me solve this. I should calculate determinant of this $n \times n$-matrix: $$ \begin{pmatrix} 1 & 9 & \cdots & 9 \\ 9 & 2 & \cdots & 9 \\ 9 & 9 & \cdots & 9 \\ \vdots & \vdots & \ddots & \vdots \\ 9 & 9 & \cdots & n \\ \end{pmatrix} $$ where $n$ is a nonnegative integer.

Answer 1

There's nothing very special about $9$, so let's consider the corresponding problem for matrices with off-diagonal elements $q$ and diagonal elements $1,2,\ldots, n$. Subtract the top row from each other row without changing the determinant: you get

$$ \pmatrix{1 & q & q & \ldots & q\cr q-1 & 2-q & 0 & \ldots & 0\cr q-1 & 0 & 3-q & \ldots & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots \cr q-1 & 0 & 0 & \ldots & n-q\cr} $$ Now use Leibniz's formula for the determinant. A term that avoids all the zero entries will be either the product $(2-q)(3-q)\ldots (n-q)$ of all the diagonal elements, or it will have one of the $q-1$'s (say in row $i$), the corresponding $q$ in column $i$, and the diagonal elements except for $1$ and $i-q$. The corresponding permutation involves a single transposition $(1,i$), so has negative signature. Thus your determinant is $$ \prod_{j=2}^n (j-q) - \sum_{i=2}^n q(q-1) \prod_{j: 2 \le j \le n, j \ne i} (j-q) $$