70 2 Algebra
⎛
⎜⎜
⎜⎜
⎜⎜
⎜
⎝1111 ··· 1 1000··· 0
1222 ··· 2 0100··· 0
1211 ··· 1 0010··· 0
1212 ... 2 0001... 0
..
.
..
.
..
.
..
.
... ..
.
..
.
..
.
..
.
..
.
... ..
.
1212 ··· ··· 0000 ··· 1
⎞
⎟⎟
⎟⎟
⎟⎟
⎟
⎠
.
Subtracting the first row from each of the others, then the second row from the first, we
obtain
⎛
⎜
⎜⎜
⎜⎜
⎜⎜
⎝
1000 ··· 02 − 100 ··· 0
0111 ··· 1 − 1100 ··· 0
0100 ··· 0 − 1010 ··· 0
0101 ··· 1 − 1001 ··· 0
..
.
..
.
..
.
..
.
... ..
.
..
.
..
.
..
.
..
.
.....
.
0101 ······ − 1000 ··· 1
⎞
⎟
⎟⎟
⎟⎟
⎟
⎟
⎠
.
We continue as follows. First, we subtract the second row from the third, fourth, and so
on. Then we add the third row to the second. Finally, we multiply all rows, beginning
with the third, by−1. This way we obtain
⎛
⎜⎜
⎜⎜
⎜⎜
⎜
⎝
1000 ··· 02 −10 0··· 0
0100 ··· 0 − 1010 ··· 0
0011 ··· 101 − 10 ··· 0
0010 ··· 0 010− 1 ··· 0
..
.
..
.
..
.
..
.
... ..
.
..
.
..
.
..
.
..
.
... ..
.
0010 ··· ··· 1000 ··· − 1
⎞
⎟⎟
⎟⎟
⎟⎟
⎟
⎠
.
Nowthe inductive pattern is clear. At each step we subtract thekth row from the rows
below, then subtract the(k+ 1 )st from thekth, and finally multiply all rows starting with
the(k+ 1 )st by−1. In the end we find that the entries ofA−^1 areb 1 , 1 =2,bn,n=(− 1 )n,
bi,i+ 1 =bi+ 1 ,i=(− 1 )i, andbij=0, for|i−j|≥2. This example shows that equality
can hold in part (a).
221.For distinct numbersx 1 ,x 2 ,...,xn, consider the matrix
A=
⎛
⎜⎜
⎜
⎝
11 ··· 1
x 1 x 2 ···xn
..
...
.
... ..
.
x 1 n−^1 x 2 n−^1 ···xnn−^1⎞
⎟⎟
⎟
⎠
.
It is known that det∏ Ais the Vandermonde determinant (x 1 ,x 2 ,...,xn) =
i>j(xi−xj).Prove that the inverse ofAisB=(bkm)^1 ≤k,m≤n, where