Explicit inverses of some tridiagonal matrices

Linear Algebra and its Applications 325 (2001) 7 21 wwwelseviercom/locate/laa Explicit inverses of some tridiagonal matrices CM da Fonseca, J Petronilho Depto de Matematica, Faculdade de Ciencias e Technologia, Univ Coimbra, Apartado 3008, 3000 Coimbra, Portugal Received 28 April 1999; accepted 26 June 2000 Submitted by G Heinig Abstract We give explicit inverses of tridiagonal 2-Toeplitz and 3-Toeplitz matrices which generalize some well-known results concerning the inverse of a tridiagonal Toeplitz matrix 2001 Elsevier Science Inc All rights reserved AMS classification: 33C45; 42C05 Keywords: Invertibility of matrices; r-toeplitz matrices; Orthogonal polynomials 1 Introduction In recent years the invertibility of nonsingular tridiagonal or block tridiagonal matrices has been quite investigated in different fields of applied linear algebra (for historical notes see [8]) Several numerical methods, more or less efficient, have risen in order to give expressions of the entries of the inverse of this kind of matrices Though, explicit inverses are known only in a few cases, in particular when the tridiagonal matrix is symmetric with constant diagonals and subject to some restrictions (cf [3,8,10]) Furthermore, Lewis [5] gave a different way to compute other explicit inverses of nonsymmetric tridiagonals matrices In [1], Gover defines a tridiagonal 2-Toeplitz matrix of order n, as a matrix A = (a ij ),a ij = 0if i j > 1anda ij = a kl if (i, j) (k, l) mod 2, ie, Corresponding author E-mail addresses: cmf@matucpt (CM da Fonseca), josep@matucpt (J Petronilho) 0024-3795/01/$ - see front matter 2001 Elsevier Science Inc All rights reserved PII:S0024-3795(00)00289-5

8 CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 a 1 b 1 c 1 a 2 b 2 c 2 a 1 b 1 A = c 1 a 2 b 2 (1) c 2 a 1 n n Similarly, a tridiagonal 3-Toeplitz matrix (n n) is of the form a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 a 1 b 1 B = c 1 a 2 b 2 (2) c 2 a 3 b 3 c 3 a 1 n n Making use of the theory of orthogonal polynomials, we will give the explicit inverse of tridiagonal 2-Toeplitz and 3-Toeplitz matrices, based on recent results from [1,6,7] As a corollary, we will obtain the explicit inverse of a tridiagonal Toeplitz matrix, ie, a tridiagonal matrix with constant diagonals (not necessarily symmetric) Different proofs of the results by Kamps [3,4] involving the sum of all the entries of the inverse can be simplified Throughout this paper, it is assumed that all the matrices are invertible 2 Inverse of a tridiagonal matrix Let T be an n n real nonsingular tridiagonal matrix a 1 b 1 0 c 1 a 2 b 2 T = c 2 (3) bn 1 0 c n 1 a n Denote T 1 = (α ij ) In [5] Lewis proved the following result Lemma 21 Let T be the matrix (3) and assume that b l /= 0 for l = 1,,n 1 Then α ji = γ ij α ij when i<j,

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 9 γ ij = j 1 l=i c l b l Of course this reduces to α ij = α ji, when T is symmetric Using this lemma, one can prove that there exist two finite sequences {u i } and {v i } (i = 1,,n 1) such that u 1 v 1 u 1 v 2 u 1 v 3 u 1 v n γ 12 u 1 v 2 u 2 v 2 u 2 v 3 u 2 v n T 1 = γ 13 u 1 v 3 γ 23 u 2 v 3 u 3 v 3 u 3 v n (4) γ 1,n u 1 v n γ 2,n u 2 v n γ 3,n u 3 v n u n v n Since {u i } and {v i } are only defined up to a multiplicative constant, we make u 1 = 1 The following result shows how to compute {u i } and {v i } The determination of these sequences is very similar to the particular case studied in [8], the reader may check for details Lemma 22 (i) The {v i } (i = 1,,n 1) can be computed from v 1 = 1 d 1, v k = b k 1 d k v k 1, k = 2,,n, d n = a n, d i = a i b ic i, i = n 1,,1 d i+1 (ii) The {u i } (i = 1,,n 1) can be computed from u n = 1, δ n v n δ 1 = a 1, u k = b k δ k u k+1, k = n 1,,1, δ i = a i b i 1c i 1 δ i 1, i = 2,,n 1 Notice that for centro-symmetric matrices, ie, a ij = a n+1 i,n+1 j, we have v i = u n+1 i The following proposition is known in the literature We give a proof based on the lecture of [8] Theorem 21 Let T be the n n tridiagonal matrix defined in (3) and assume that b l /= 0 for l = 1,,n 1Then

10 CM 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 jthen 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

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 11 3 Chebyshev polynomials of the second kind In the next it is useful to consider the set of polynomials {U n } n 0, such that each U n is of degree exactly n, satisfying the recurrence relation U n+1 (x) = 2xU n (x) U n 1 (x), n 1 with initial conditions U 0 = 1 and U 1 = 2x These polynomials are called Chebyshev polynomials of the second kind and they have the form sin(n + 1)θ U n (x) =, cos θ = x, sin θ when x < 1 When x > 1, they have the form sinh(n + 1)θ U n (x) =, cosh θ = x sinh θ In particular U n (±1) = (±1) n (n + 1) In any case, if x /= 1, then U n (x) = rn+1 + rn+1, r + r r ± = x ± x 2 1 are the two solutions of the quadratic equation r 2 2xr + 1 = 0 The following proposition is a slight extension of Lemma 25 of Meurant [8] Lemma 31 Fix a positive integer number n and let a and b be nonzero real numbers such that U i (a/2b) /= 0 for all i = 1, 2,,n Consider the recurrence relation α 1 = a, α i = a b2, i = 2,,n α i 1 Under these conditions, if a =±2b, then α i = Otherwise ±b(i + 1) i α i = b ri+1 r+ i ri + ri+1,

12 CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 r ± = a ± a 2 4b 2 2b are the two solutions of the quadratic equation r 2 (a/b)r + 1 = 0 Proof Let us assume that a/=±2b (otherwise it is trivial) and set α i = β i /β i 1 Then β i aβ i 1 + b 2 β i 2 = 0, β 0 = 1, β 1 = a, hence β i = b i U i (a/2b) and the conclusion follows 4 Inverse of a tridiagonal 2-Toeplitz matrix In this section we will determine the explicit inverse of a matrix of type (1) By Theorem 21, this is equivalent to determine δ i and d i, ie, by (7) and transformations (5) and (6), θ i and φ i So, to evaluate θ i,wehave θ 0 = 1, θ 1 = a 1, θ 2i = a 2 θ 2i 1 b 1 c 1 θ 2i 2, θ 2i+1 = a 1 θ 2i b 2 c 2 θ 2i 1 Making the change z i = ( 1) i θ i, we get the relations z 0 = 1, z 1 = a 1, z 2i = a 2 z 2i 1 b 1 c 1 z 2i 2, z 2i+1 = a 1 z 2i b 2 c 2 z 2i 1 According to the results in [6], z i = Q i (0), Q i is a polynomial of degree exactly i, defined by the recurrence relation Q i+1 (x) = (x β i )Q i (x) γ i Q i 1 (x), (8) { β 2j = a 1, β 2j+1 = a 2, and { γ2j = b 2 c 2, γ 2j+1 = b 1 c 1, (9) and initial conditions Q 0 (x) = 1andQ 1 (x) = x β 0

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 13 The solution of the recurrence relation (8) with coefficients (9) is Q 2i (x) = P i [π 2 (x)], Q 2i+1 (x) = (x a 1 )Pi [π 2 (x)], (10) π 2 (x) = (x a 1 )(x a 2 ), (11) ( ) i x Pi (x) = b1 c 1 b 2 c 2 b1 b 2 c 1 c 2 Ui 2, (12) b 1 b 2 c 1 c 2 and ( ) [ i x b1 c 1 b 2 c 2 P i (x)= b1 b 2 c 1 c 2 U i 2 b 1 b 2 c 1 c ] 2 x b1 c 1 b 2 c 2 + βu i 1 2, (13) b 1 b 2 c 1 c 2 with β 2 = b 2 c 2 /b 1 c 1 We may conclude that θ i = ( 1) i Q i (0) Now, to evaluate φ i we put ψ i = φ n+1 i, for i = 1,,n, and we have to distinguish the cases when the order n of the matrix (1) is odd and when it is even First let us suppose that n is odd Then we have ψ 0 = 1, ψ 1 = a 1, ψ 2i = a 2 ψ 2i 1 b 2 c 2 ψ 2i 2, ψ 2i+1 = a 1 ψ 2i b 1 c 1 ψ 2i 1 Following the same steps as in the previous case, we obtain ψ i = ( 1) i Q i (0), ie, φ i = ( 1) i Q n+1 i (0), Q i (x) is the same as (10) but with β 2 = b 1 c 1 /b 2 c 2 in (13) Notice that if b 1 c 1 = b 2 c 2,thenθ i = φ n+1 i If n is even, we get ψ 0 = 1, ψ 1 = a 2, ψ 2i = a 1 ψ 2i 1 b 1 c 1 ψ 2i 2, ψ 2i+1 = a 2 ψ 2i b 2 c 2 ψ 2i 1 Therefore ψ i = ( 1) i Q i (0),

14 CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 ie, φ i = ( 1) i+1 Q n+1 i (0), Q 2i (x) = P i [π 2 (x)], Q 2i+1 (x) = (x a 2 )Pi [π 2 (x)], with π 2 (x), P i (x) and Pi (x) the same as in (11), (13) and (12), respectively Observe that if a 1 = a 2, then θ i = φ n+1 i We have determined completely the θ i s and φ i s of (7 ) thus the inverse of the matrix in the case of a tridiagonal 2-Toeplitz matrix (1) Theorem 41 Let A be the tridiagonal matrix (1), with a 1 a 2 /= 0 and b 1 b 2 c 1 c 2 > 0 Put π 2 (x):=(x a 1 )(x a 2 ), β := b 2 c 2 /b 1 c 1 and let {Q i ( ; α, γ )} be the sequence of polynomials defined by ( ) [ i π2 (x) b 1 c 1 b 2 c 2 Q 2i (x; α, γ )= b1 b 2 c 1 c 2 U i 2 b 1 b 2 c 1 c ] 2 π2 (x) b 1 c 1 b 2 c 2 +γu i 1 2 b 1 b 2 c 1 c 2 ( ) i π2 (x) b 1 c 1 b 2 c 2 Q 2i+1 (x; α, γ ) = (x α) b1 b 2 c 1 c 2 Ui 2, b 1 b 2 c 1 c 2 α and γ are some parameters Under these conditions, ( 1) i+j b (A 1 p (j i)/2 i bq (j i+1)/2 i θ i 1 φ j+1 /θ n if i j, ) ij = ( 1) i+j c (j i)/2 cq (j i+1)/2 j θ j 1 φ i+1 /θ n if i>j, p j (14) p l = (3 ( 1) l )/2, q l = (3 + ( 1) l )/2, z denotes the greater integer less or equal to the real number z, θ i = ( 1) i Q i (0; a 1,β), (15) and ( 1) i Q n+1 i (0; a 1, 1/β) φ i = ( 1) i+1 Q n+1 i (0; a 2,β) if n is odd, if n is even Remark As we already noticed, under the conditions of Theorem 41, if n is odd and b 1 c 1 = b 2 c 2, then θ i = φ n+1 i ;andifn is even and a 1 = a 2, then also θ i = φ n+1 i

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 15 As a first application, let us consider a classical problem on the inverse of a tridiagonal Toeplitz matrix Suppose that a 1 = a 2 = a, b 1 = b 2 = b and c 1 = c 2 = c in (1), with a/= 0andbc > 0 Therefore, we have the n n tridiagonal Toeplitz matrix a b 0 c a b T = c a (16) b 0 c a According to (14), ( 1) i+j b j i θ i 1 φ j+1 /θ n if i j, (T 1 ) ij = ( 1) i+j c i j θ j 1 φ i+1 /θ n if i>j It is well-known and we can easily check using elementary trigonometry that ) U 2i+1 (x) = 2xU i (2x 2 1 Since β = 1and ( a 2 ) ( 2bc U i = U i 2bc from (15) we get ( ) ( i a θ i = bc Ui a 2 2 2 1) = bc bc a a U 2i+1 2, bc ) 2, bc and taking into account the above remark the φ i s can be computed using the θ i s, giving ( ) n+1 i a φ i = θ n+1 i = bc Un+1 i 2 bc Therefore, the next result comes immediately Corollary 41 Let T be the matrix in (16) and define d :=a/2 bc The inverse is given by ( 1) i+j b j i U i 1 (d)u n j (d) ( bc ) j i+1 if i j, U n (d) (T 1 ) ij = ( 1) i+j c i j U j 1 (d)u n i (d) ( bc ) i j+1 if i>j U n (d)

16 CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 This result is well-known It can be deduced, eg, using results in the book of Heinig and Rost [2, p 28] The next corollary is Theorem 31 of Kamps [3] Corollary 42 Let Σ be the n-square matrix a b 0 b a b Σ = b a (17) b 0 b a The inverse is given by ( 1) i+j 1 b (Σ 1 ) ij = ( 1) i+j 1 b U i 1 (a/2b)u n j (a/2b) U n (a/2b) U j 1 (a/2b)u n i (a/2b) U n (a/2b) if i j, if i>j We consider now a second application In [3,4], the matrices of type (17), with a>0, b/= 0anda>2 b, arose as the covariance matrix of one-dependent random variables Y 1,,Y n, with same expectation Let us consider the least squares estimator ˆµ opt = 1t Σ 1 Y 1 t Σ 1 1, 1 = (1,,1) and Y = (Y 1,,Y n ) t, which estimates the parameter µ equal to the common expectation of the Y i s, with variance V 1 ˆµ opt = 1 t Σ 1 1 According to Kamps, the estimator ˆµ opt is the best unbiased estimator based on Y So, the sum of all the entries of the inverse of Σ, 1 t Σ 1 1, has an important role in the determination of this estimator and therefore in the computation of the variance V ( ˆµ opt ) Kamps used in [3] some sum and product formulas involving different kinds of Chebyshev polynomials We will prove the same results in a more concise way Corollary 43 The sum s i of the ith row (or column) of Σ 1, for i = 1,,n is given by s i = 1 + b(σ 1i + σ 1,n i+1 ), a + 2b σ ij = (Σ 1 ) ij, when i j Proof By Corollary 42, for i j,

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 17 ( σ ij = ( 1) i+j 1 r i + r ) i r n j+1 + r n j+1 ( ), b (r + r ) r+ n+1 rn+1 r ± = a ± a 2 4b 2 2b Thus i 1 n s i = σ ji + σ ij j=1 j=1 j=i By the sum of a geometric progression and the fact that r + r = 1, we have i 1 ( r+ i+1 )( and σ ji = b (σ 1i + σ ii ) a + 2b + 1 a + 2b n σ ij = b σ 1,n i+1 σ ii a + 2b j=i 1 a + 2b (r + r ) ( r i + r i ) ( r n i ) r+ n i+1 r n i+1 ( ) r+ n+1 rn+1 + rn i ( (r + r ) r+ n+1 rn+1 ) ) Finally ( r+ i+1 ri+1 = (r + r ) r+ n i+1 r n i+1 ( r+ n+1 rn+1 ) r+ i ri r+ n i rn i ) The previous theorem says that s i depends only on the first row of Σ 1 In fact (as the following result says) the sum of all entries of Σ 1 depends only on σ 11 and σ 1n Corollary 44 1 t Σ 1 1 = n + 2bs 1 a + 2b 5 Inverse of a tridiagonal 3-Toeplitz matrix In an analogous way of the 2-Toeplitz matrices, we may ask about the inverse of a tridiagonal 3-Toeplitz matrix To solve this case, we have to compute θ i and φ i in (7)

18 CM 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), 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), β 3j = a 1, γ 3j = b 3 c 3, β 3j+1 = a 2, and γ 3j+1 = b 1 c 1, β 3j+2 = a 3, 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)], 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

CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 19 We may conclude that θ i = ( 1) i Q i (0) In order to compute the φ i s, we define ψ i = φ n+1 i for i = 1,,n In this situation, we have to distinguish three cases If n 0 (mod 3), then ψ 0 = 1, ψ 1 = a 3, ψ 3i = a 1 ψ 3i 1 b 1 c 1 ψ 3i 2, ψ 3i+1 = a 3 ψ 3i b 3 c 3 ψ 3i 1, ψ 3i+2 = a 2 ψ 3i+1 b 2 c 2 ψ 3i If n 1(mod3),then ψ 0 = 1, ψ 1 = a 1, ψ 3i = a 2 ψ 3i 1 b 2 c 2 ψ 3i 2, ψ 3i+1 = a 1 ψ 3i b 1 c 1 ψ 3i 1, ψ 3i+2 = a 3 ψ 3i+1 b 3 c 3 ψ 3i, and if n 2(mod3),then ψ 0 = 1, ψ 1 = a 2, ψ 3i = a 3 ψ 3i 1 b 3 c 3 ψ 3i 2, ψ 3i+1 = a 2 ψ 3i b 2 c 2 ψ 3i 1, ψ 3i+2 = a 1 ψ 3i+1 b 1 c 1 ψ 3i It is clear that these three recurrence relations can be solved by a similar process as for the computation of the θ i s (mutatis mutandis) Following in this way, we can state the next proposition Theorem 51 Let B be the tridiagonal matrix (2), with a 1 a 2 a 3 /= 0 and b 1 b 2 b 3 c 1 c 2 c 3 > 0Let π 3 α β γ ; x x a 1 1 := α x b 1 γ β x c, P i α β γ ; x :=2 αβγ ( [ ]) 1 U i 2 π 3 αβγ α β γ ; x (a + b + c),

20 CM da Fonseca, J Petronilho / Linear Algebra and its Applications 325 (2001) 7 21 Q 3i α β γ ; x := P i α β γ ; x +γ(x b) P i 1 α β γ ; x, Q 3i+1 α β γ ; x := (x a)p i α β γ ; x +αγ P i 1 α β γ ; x, Q 3i+2 α β γ ; x := [(x a)(x b) α] P i α β γ ; x, a, b, c, α, β and γ are some parameters, subject to the restriction αβγ > 0 Under these conditions, ( 1) i+j β i β i+1 β j θ i 1 φ j+1 /θ n if i j, (B 1 ) ij = (19) ( 1) i+j γ j γ j+1 γ i θ j 1 φ i+1 /θ n if i>j, β 3l+s+1 = b s+1, γ 3l+s+1 = c s+1 (s = 0, 1, 2; l = 0, 1, 2,), and the θ i s and φ i s are explicitly given by θ i = ( 1) i a1 a Q 2 a 3 i ; 0 (20) b 1 c 1 b 2 c 2 b 3 c 3 and ( 1) n+1 i a3 a Q 2 a 1 n+1 i ; 0 if n 0, (mod 3), b 2 c 2 b 1 c 1 b 3 c 3 φ i = ( 1) n+1 i a1 a Q 3 a 2 n+1 i ; 0 b 3 c 3 b 2 c 2 b 1 c 1 ( 1) n+1 i a2 a Q 1 a 3 n+1 i ; 0 b 1 c 1 b 3 c 3 b 2 c 2 if n 1, (mod 3), if n 2,(mod 3) Acknowledgments This work was supported by CMUC (Centro de Matemática da Universidade de Coimbra) The authors thank the editor as well as the unknown referees for their helpful remarks which helped to improve the final version of the paper

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