Explicit inverses of some tridiagonal matrices - Estudo Geral ...

18 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
For θ i ,wehave
θ 0 = 1, θ 1 = a 1 ,
θ 3i = a 3 θ 3i−1 − b 2 c 2 θ 3i−2 ,
θ 3i+1 = a 1 θ 3i − b 3 c 3 θ 3i−1 ,
θ 3i+2 = a 2 θ 3i+1 − b 1 c 1 θ 3i .
Making the change
z i = (−1) i θ i ,
we get the relations
z 0 = 1, z 1 =−a 1 ,
z 3i =−a 3 z 3i−1 − b 2 c 2 z 3i−2 ,
z 3i+1 =−a 1 z 3i − b 3 c 3 z 3i−1 ,
z 3i+2 =−a 2 z 3i+1 − b 1 c 1 z 3i .
According to the results in [7],
z i = Q i (0),
where Q i is a polynomial of degree exactly i, defined according to
Q i+1 (x) = (x − ˜β i )Q i (x) −˜γ i Q i−1 (x),
where
⎧⎨
⎩
˜β 3j = a 1 ,
˜β 3j+1 = a 2 ,
˜β 3j+2 = a 3 ,
and
⎧
⎨ ˜γ 3j = b 3 c 3 ,
˜γ 3j+1 = b 1 c 1 ,
⎩
˜γ 3j+2 = b 2 c 2 .
The solution of the recurrence relation (8) with coefficients (18) is
(18)
Q 3i (x) = P i [π 3 (x)] + b 3 c 3 (x − a 2 ) P i−1 [π 3 (x)] ,
Q 3i+1 (x) = (x − a 1 )P i [π 3 (x)] + b 1 c 1 b 3 c 3 P i−1 [π 3 (x)] ,
Q 3i+2 (x) = [(x − a 1 )(x − a 2 ) − b 1 c 1 ] P i [π 3 (x)] ,
where
∣ x − a 1 1 1 ∣∣∣∣∣
π 3 (x) =
b 1 c 1 x − a 2 1
∣ b 3 c 3 b 2 c 2 x − a 3
and
P i (x) =
(2 √ ) [ ( )]
i x − b1 c 1 − b 2 c 2 − b 3 c 3
b 1 b 2 b 3 c 1 c 2 c 3 U i
2 √ .
b 1 b 2 b 3 c 1 c 2 c 3