Distances along a line

Idea. Neighbouring rows differ by $\pm 1$ in every entry, and twice-differenced rows are almost all zero. Two rounds of row subtractions leave a matrix that is easy to expand.

Step 1: first differences. Replace row $i$ by row $i$ minus row $i+1$, for $i=1,2,3,4$ in that order (each time the row below is still the original one). The determinant does not change, and the rows become $$\begin{array}{rl}& (-1,1,1,1,1),(-1,-1,1,1,1),\\ & (-1,-1,-1,1,1),(-1,-1,-1,-1,1),\\ & (4,3,2,1,0).\end{array}$$

Step 2: second differences. Do the same again for rows 1, 2, 3: $$\begin{array}{rl}& (0,2,0,0,0),(0,0,2,0,0),(0,0,0,2,0),\\ & (-1,-1,-1,-1,1),(4,3,2,1,0).\end{array}$$

Step 3: expand. Expand along the first three rows, each with a single entry 2, in columns 2, 3, 4. Each expansion contributes a factor $2$ with sign $-1$ (the entry sits one place to the right of the diagonal of the remaining matrix), so together $(-2{)}^3=-8$ times the determinant of what is left in columns 1 and 5: $$\det \left(\begin{array}{cc}-1& 1\\ 4& 0\end{array}\right)=-4.$$ So $\det A=(-8)(-4)=32$.

In general. For the $n\times n$ matrix $|i-j|$, $$\det A=(-1{)}^{n-1}(n-1)2^{n-2}.$$ For $n=5$ this gives $4\cdot 8=32$; for $n=4$ it gives $-12$.