Question states: "Let $D_n$ denote the determinant of the $n$-by-$n$ matrix whose $(i,j)$th element (the element in the $i$th row and $j$th column) is the absolute value of the difference between $i$ and $j$. Show that $D_n = (-1)^{n-1}(n-1)2^{n-2}$."
It seems like that I can get a recurrence relation (especially the question was from a book chapter dealing with recursion). In trying to get a recurrence relation, here is what I have done so far:
(Let $R_k$ denote the $k$th row and let $(i,j)$ denote the element/spot of $i$th row and $j$th spot)
1) replace $R_1$ with $R_1-(R_2+R_{n-1})$ to get $R_1^*$
2) replace $R_1^*$ with $R_1^*+R_{n-2}$
3) If I am not mistaken, I should have $-2$ on $(1,1)$, $2$ on both $(1,n-1)$ and $(1,n)$, and all other element in the first row as $0$.
4) take out $2$ from the first row, and expand the determinant about the first row
5) I should have $D_n=2(-D_{n-1}+(-1)^n B_{n-1})$, where $B_{n-1}$ is the determinant of a not-easily-reducible mayrix whose last column consists of $n-1$ ones followed by one $-1$.
How should I go about from now on? Once I get a recurrence relation, it would be pretty easy to get the explicit formula for $D_n$ using induction.
P.S. This question apparently is a question taken from 1969 Putnam Competition
As I said in the comments, the solution is here. I am copying it on this page just in case the referencing page gets deleted.
For $i = 1, 2, 3, ... , n-2$ subtract twice row $i+1$ from row $i$ and add row $i+2$ to row $i$. For $i < n-1$, row $i$ becomes all $0$s except for a $2$ in column $i+1$.
Now expand successively by the first, second, third ... rows to get $(-2)n-2$ times the $2 \times 2$ determinant with first row $n-2, 1$ and second row $n-1, 0$. This $2 \times 2$ determinant has value $-(n-1)$, so $|A| = (-1)^{n-1} (n-1) 2^{n-2}$.