110 4. Particular Determinants
Proof. The identity is a particular case of Jacobi variant (A) (Sec-
tion 3.6.3),
∣
∣
∣
∣
∣
T
(n)
ip
T
(n+1)
i,n+1
T
(n)
jp
T
(n+1)
j,n+1
∣
∣
∣
∣
∣
−TnT
(n+1)
ij;p,n+1=0, (4.8.20)
where (i, j, p)=(1,n,1).
Let
An=T
(n,r)
,
Bn=T
(n,r+1)
.
Then Theorem 4.29 is satisfied by bothAnandBn.
Theorem 4.30. For all values ofr,
a.AnB
(n+1)
n+1,n
−BnA
(n+1)
n+1,n
+An+1Bn− 1 =0.
b.Bn− 1 A
(n+1)
n+1,n−AnB
(n)
n,n− 1 +An−^1 Bn=0.
Proof.
Bn=(−1)
n
A
(n+1)
1 ,n+1
,
B
(n+1)
n+1,n=(−1)
n
A
(n+1)
n 1 ,
Bn− 1 =(−1)
n− 1
A
(n)
1 n
=(−1)
n
A
(n+1)
n,n+1;1,n+1,
A
(n+1)
1 n;n,n+1
=A
(n)
1 ,n− 1
=(−1)
n− 1
B
(n)
n,n− 1
. (4.8.21)
Denote the left-hand side of (a) byYn. Then, applying the Jacobi identity
toAn+1,
(−1)
n
Yn=
∣
∣
∣
∣
∣
A
(n+1)
n 1
A
(n+1)
n,n+1
A
(n+1)
n+1, 1
A
(n+1)
n+1,n+1
∣
∣
∣
∣
∣
−An+1A
(n+1)
n,n+1;1,n+1
=0,
which proves (a).
The particular case of (4.8.20) in which (i, j, p)=(n, 1 ,n) andT is
replaced byAis
∣
∣
∣
∣
∣
An− 1 A
(n+1)
n,n+1
A
(n)
1 n A
(n+1)
1 ,n+1
∣
∣
∣
∣
∣
−AnA
(n+1)
n1;n,n+1
=0. (4.8.22)
The application of (4.8.21) yields (b).
This theorem is applied in Section 6.5.1 on Toda equations.