Page 155 - Determinants and Their Applications in Mathematical Physics

140 4. Particular Determinants

−1 2
= K n K
n Q n − t I n
2 2

= K n S − t I n
n
= K n (S n + tI n )(S n − tI n )
= K n H n H n .
Corollary.
B −1 = H −1 H −1 K −1 ,
n n n n
(n) (n) (n) (n)
B = H H K .
ji ji ji ji
Lemma.
n
(n) 2j−1
h = x + t.
ij
i=1
The proof applies (4.11.3) and is elementary.
Let E n+1 denote the determinant of order (n + 1) obtained by bordering
H n as follows:
v n1 /n
h 11 h 12 ··· h 1n

h 21 h 22 ··· h 2n v n2 /(n +1)

E n+1 = ..................................

h n1 h n2 ··· h nn v nn /(2n − 1)
1 1 ··· 1 •

n+1
n n

= − v nr H rs . (4.11.16)
n + r − 1
r=1 s=1
Theorem 4.45.
E n+1 =(−1) H n−1 .
n
The proof consists of a sequence of row and column operations.
Proof. Perform the column operation

C = C n − x 2n−1 C n+1 (4.11.17)
n
and apply (6b) with j = n. The result is

h 11 h 12 ··· h 1,n−1 • v n1 /n

h 21 h 22 ··· h 2,n−1 • v n2 /(n +1)
. (4.11.18)

E n+1 = ........................................

h n1 h n2 ··· h n,n−1 t v nn /(2n − 1)
1 1 ··· 1 1 •

n+1
Remove the element in position (n, n) by performing the row operation
R = R n − tR n+1 . (4.11.19)

n