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

10 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
⎧
(−1) ⎪⎨
i+j d j+1 ···d n
b i ···b j−1 if i j,
δ i ···δ n
(T −1 ) ij =
⎪⎩ (−1) i+j d i+1 ···d n
c j ···c i−1 if i>j
δ j ···δ n
(with the convention that empty product equals 1).
Proof. Let us assume that i j.Then
But
(T −1 ) ij = u i v j = (−1) n−i b i ···b n−1
δ i ···δ n v n
× (−1) j−1 b 1 ···b j−1
d 1 ···d j
.
v n = (−1) n−1 b 1 ···b n−1
.
d 1 ···d n
Therefore
(T −1 ) ij =(−1) n−i b i ···b n−1
δ i ···δ n
× (−1) n−1 d 1 ···d n
b 1 ···b n−1
× (−1) j−1 b 1 ···b j−1
d 1 ···d j
=(−1) i+j b i ···b j−1
d j+1 ···d n
δ i ···δ n
. □
If we set
δ i =
θ i
,
θ i−1
θ 0 = 1, θ 1 = a 1 , (5)
we get the recurrence relation
θ i = a i θ i−1 − b i−1 c i−1 θ i−2 ;
and if we now put
d i =
φ i
, φ n+1 = 1, φ n = a n , (6)
φ i+1
we get the recurrence relation
φ i = a i φ i+1 − b i c i φ i+2 .
As a consequence, we will achieve
⎧
⎨(−1) i+j b i ···b j−1 θ i−1 φ j+1 /θ n if i j,
(T −1 ) ij =
(7)
⎩
(−1) i+j c j ···c i−1 θ j−1 φ i+1 /θ n if i>j.
These are the general relations given by Usmani [9], which we will use throghout
this paper. Notice that
θ n = δ 1 ···δ n = det T.