Algebra 417The Vandermonde determinant has the value
∏
i>j(xi−xj).
It follows that our determinant is equal to∏
i>j(xi−xj)/(^1!^2 !···(n−^1 )!), which
therefore must be an integer. Hence the conclusion.
(Mathematical Mayhem, 1995)
210.The determinant is annth-degree polynomial in each of thexi’s. (If you have a
problem working with multinomials, think ofx 1 as the variable and of the others as
parameters!) Adding all other columns to the first, we obtain that the determinant is
equal to zero whenx 1 +x 2 + ··· +xn =0, sox 1 +x 2 + ··· +xnis a factor of the
polynomial. This factor corresponds toj=0 on the right-hand side of the identity from
the statement. For some otherj, multiply the first column byζj, the second byζ^2 j, and
so forth; then add all columns to the first. As before, we see that the determinant is zero
when
∑n
k= 1 ζ
jkxk=0, so∑n
k= 1 ζ
jkxkis a factor of the determinant. No two of thesepolynomials are a constant multiple of the other, so the determinant is a multiple of
n∏− 1j= 1(n
∑k= 1ζjkxk)
.
The quotient of the two is a scalarC, independent ofx 1 ,x 2 ,...,xn. Forx 1 =1,x 2 =
x 3 = ··· =xn=0, we obtain
x 1 n=Cn∏− 1j= 1(ζjx 1 )=Cζ^1 +^2 +···+(n−^1 )x 1 =Cζn(n−^1 )/^2 x 1 n=Ce(n−^1 )π ixn 1 =C(− 1 )n−^1 x 1.HenceC=(− 1 )n−^1 , which gives rise to the formula from the statement.
211.By adding the second row to the first, the third row to the second,...,thenth row
to the(n− 1 )st, the determinant does not change. Hence
det(A)=∣
∣∣
∣
∣∣
∣∣
∣∣
∣∣
∣
2 − 1 + 1 ··· ± 1 ∓ 1
− 12 − 1 ··· ∓ 1 ± 1
+ 1 − 12 ··· ± 1 ∓ 1
..
.
..
.
..
.
... ..
.
..
.
∓ 1 ± 1 ∓ 1 ··· 2 − 1
± 1 ∓ 1 ± 1 ··· − 12
∣
∣∣
∣∣
∣
∣∣
∣∣
∣∣
∣
=
∣
∣∣
∣∣
∣
∣∣
∣∣
∣∣
∣
1100 ··· 00
0110 ··· 00
0011 ··· 00
..
.
..
.
..
.
..
.
... ..
.
..
.
0000 ... 11
± 1 ∓ 1 ± 1 ∓ 1 ...− 12
∣
∣∣
∣∣
∣
∣∣
∣∣
∣∣
∣
.
Nowsubtract the first column from the second, then subtract the resulting column from
the third, and so on. This way we obtain