2.3 Linear Algebra 65Solution.We have
∣
∣∣
∣∣
∣∣
∣∣
∣
∣∣
∣
11 ··· 11
11 ··· 11
a 11 a 12 ··· a 1 ,n− 1 a 1 n
a 21 a 22 ··· a 2 ,n− 1 a 2 n
..
...
.
... ..
.
..
.
an− 2 , 1 an− 2 , 2 ···an− 2 ,n− 1 an− 2 ,n∣
∣∣
∣∣
∣∣
∣∣
∣
∣∣
∣
= 0.
Expanding by the first row, we obtain
A 1 −A 2 +A 3 −A 4 +··· = 0.This implies
A 1 +A 3 +A 5 +··· =A 2 +A 4 +A 6 +···,and (a) is proved.
For (b), we substituteaij=xij,i= 1 , 2 ,...,n−2,j= 1 , 2 ,...,n. Then
Ak=∣
∣∣
∣∣
∣
∣∣
∣
1 ··· 11 ... 1
x 1 ···xk− 1 xk+ 1 ··· xn
..
.... ..
.
..
.
... ..
.
x 1 n−^2 ···xkn−− 12 xnk+−^21 ···xnn−^2∣
∣∣
∣∣
∣
∣∣
∣
,
which is a Vandermonde determinant. Its value is equal to
∏i>j
i,j =k(xi−xj)=1
pkqk.
The equality proved in (a) becomes, in this particular case,
∑nk= 1(− 1 )k
pkqk= 0 ,
as desired.
Finally, if in this we letxk=k^2 , then we obtain the identity from part (c), and the
problem is solved.
And here comes a set of problems for the reader.