The Invertibility, Explicit Determinants, and Inverses of Circulant and Left Circulant and 𝑔-Circulant Matrices Involving Any Continuous Fibonacci and Lucas Numbers

Abstract

Circulant matrices play an important role in solving delay differential equations. In this paper, circulant type matrices including the circulant and left circulant and 𝑔-circulant matrices with any continuous Fibonacci and Lucas numbers are considered. Firstly, the invertibility of the circulant matrix is discussed and the explicit determinant and the inverse matrices by constructing the transformation matrices are presented. Furthermore, the invertibility of the left circulant and 𝑔-circulant matrices is also studied. We obtain the explicit determinants and the inverse matrices of the left circulant and 𝑔-circulant matrices by utilizing the relationship between left circulant, 𝑔-circulant matrices and circulant matrix, respectively.

1. Introduction

Circulant matrices have important applications in solving various differential equations [1–3]. The use of circulant preconditioners for solving structured linear systems has been studied extensively since 1986; see [4, 5]. Circulant matrices also play an important role in solving delay differential equations. In [6], Chan et al. proposed a preconditioner called the Strang-type block-circulant preconditioner for solving linear systems from IVPs. The Strang-type preconditioner was also used to solve linear systems from differential-algebraic equations and delay differential equations; see [7–14]. In [15], Jin et al. proposed the GMRES method with the Strang-type block-circulant preconditioner for solving singular perturbation delay differential equations.

The 𝑔-circulant matrices play an important role in various applications as well; please refer to [16, 17] for details. There are discussions about the convergence in probability and in distribution of the spectral norm of 𝑔-circulant matrices in [18, 19]. Ngondiep et al. showed the singular values of 𝑔-circulants in [20].

Recently, some scholars have given various algorithms for the determinants and inverses of nonsingular circulant matrices [21, 22]. Unfortunately, the computational complexity of these algorithms is increasing dramatically with the increasing order of matrices. However, some authors gave the explicit determinants and inverse of circulant involving Fibonacci and Lucas numbers. For example, Jaiswal evaluated some determinants of circulant whose elements are the generalized Fibonacci numbers [23]. Lind presented the determinants of circulant involving Fibonacci numbers [24]. Lin gave the determinant of the Fibonacci-Lucas quasicyclic matrices in [25]. Shen et al. considered circulant matrices with Fibonacci and Lucas numbers and presented their explicit determinants and inverses [26]. Bozkurt and Tam gave determinants and inverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers in [27].

The purpose of this paper is to obtain the explicit determinants, explicit inverses of circulant, left circulant, and 𝑔-circulant matrices involving any continuous Fibonacci numbers and Lucas numbers. And we generalize the result in [26].

In the following, let 𝑟 be a nonnegative integer. We adopt the following two conventions 0⁰=1, and for any sequence {𝑎_𝑛},∑^𝑛_𝑘=𝑖𝑎_𝑘=0 in the case 𝑖>𝑛.

The Fibonacci and Lucas sequences are defined by the following recurrence relations [23–26], respectively: 𝐹_𝑛+1=𝐹_𝑛+𝐹_𝑛−1where𝐹₀=0,𝐹₁𝐿=1,_𝑛+1=𝐿_𝑛+𝐿_𝑛−1where𝐿₀=2,𝐿₁=1, (1) for 𝑛≥0. The first few values of the sequences are given by the following table: 𝑛012345678⋯𝐹_𝑛𝐿01123581321⋯_𝑛.2134711182947⋯ (2) Let 𝛼 and 𝛽 be the roots of the characteristic equation 𝑥²−𝑥−1=0; then the Binet formulas of the sequences {𝐹_𝑟+𝑛} and {𝐿_𝑟+𝑛} have the form 𝐹_𝑟+𝑛=𝛼^𝑟+𝑛−𝛽^𝑟+𝑛𝛼−𝛽,𝐿_𝑟+𝑛=𝛼^𝑟+𝑛+𝛽^𝑟+𝑛. (3)

Definition 1 (see [21, 22]). In a right circulant matrix (or simply, circulant matrix) Circ𝑎₁,𝑎₂,…,𝑎_𝑛=𝑎₁𝑎₂⋯𝑎_𝑛𝑎_𝑛𝑎₁⋯𝑎_𝑛−1𝑎⋮⋮⋮₂𝑎₃⋯𝑎₁, (4) each row is a cyclic shift of the row above to the right. Right circulant matrix is a special case of a Toeplitz matrix. It is evidently determined by its first row (or column).

Definition 2 (see [22, 28]). In a left circulant matrix (or reverse circulant matrix ) LCirc𝑎₁,𝑎₂,…,𝑎_𝑛=𝑎₁𝑎₂⋯𝑎_𝑛𝑎₂𝑎₃⋯𝑎₁𝑎⋮⋮⋮_𝑛𝑎₁⋯𝑎_𝑛−1, (5) each row is a cyclic shift of the row above to the left. Left circulant matrix is a special Hankel matrix.

Definition 3 (see [19, 29]). A 𝑔-circulant matrix is an 𝑛×𝑛 complex matrix with the following form: 𝐴_𝑔,𝑛=𝑎₁𝑎₂⋯𝑎_𝑛𝑎_{𝑛−𝑔+1}𝑎_{𝑛−𝑔+2}⋯𝑎_𝑛−𝑔𝑎_{𝑛−2𝑔+1}𝑎_{𝑛−2𝑔+2}⋯𝑎_𝑛−2𝑔𝑎⋮⋮⋱⋮_𝑔+1𝑎_𝑔+2⋯𝑎_𝑔, (6) where 𝑔 is a nonnegative integer and each of the subscripts is understood to be reduced modulo 𝑛.

The first row of 𝐴_𝑔,𝑛 is (𝑎₁,𝑎₂,…,𝑎_𝑛); its (𝑗+1)th row is obtained by giving its 𝑗th row a right circular shift by 𝑔 positions (equivalently, 𝑔 mod 𝑛 positions). Note that 𝑔=1 or 𝑔=𝑛+1 yields the standard circulant matrix. If 𝑔=𝑛−1, then we obtain the so called left circulant matrix.

Lemma 4 (see [26]). Let 𝐴=Circ(𝑎₁,𝑎₂,…,𝑎_𝑛) be circulant matrix; then one has(i)𝐴 is invertible if and only if the eigenvalues of 𝐴𝜆_𝑘=𝑓𝜔^𝑘≠0,(𝑘=0,1,…,𝑛−1), (7)where 𝑓(𝑥)=∑^𝑛_𝑗=1𝑎_𝑗𝑥^𝑗−1 and 𝜔=exp(2𝜋𝑖/𝑛);(ii)if 𝐴 is invertible, then the inverse 𝐴⁻¹ of 𝐴 is a circulant matrix.

Lemma 5. Define Δ∶=;100⋯00000⋯01000⋯10⋮⋮⋱⋮001⋯00010⋯00 (8) the matrix Δ is an orthogonal cyclic shift matrix (and a left circulant matrix). It holds that LCirc(𝑎₁,𝑎₂,…,𝑎_𝑛)=ΔCirc(𝑎₁,𝑎₂,…,𝑎_𝑛).

Lemma 6 (see [29]). The 𝑛×𝑛 matrix ℚ_𝑔 is unitary if and only if (𝑛,𝑔)=1, where ℚ_𝑔 is a 𝑔-circulant matrix with first row 𝑒^∗=[1,0,…,0].

Lemma 7 (see [29]). 𝐴_𝑔,𝑛 is a 𝑔-circulant matrix with first row [𝑎₁,𝑎₂,…,𝑎_𝑛] if and only if 𝐴_𝑔,𝑛=ℚ_𝑔𝐶, where 𝐶=Circ(𝑎₁,𝑎₂,…,𝑎_𝑛).

2. Determinant, Invertibility, and Inverse of Circulant Matrix with Any Continuous Fibonacci Numbers

In this section, let 𝐴_𝑟,𝑛=Circ(𝐹_𝑟+1,𝐹_𝑟+2,…,𝐹_𝑟+𝑛) be a circulant matrix. Firstly, we give the determinant equation of the matrix 𝐴_𝑟,𝑛. Afterwards, we prove that 𝐴_𝑟,𝑛 is an invertible matrix for 𝑛>2, and then we find the inverse of the matrix 𝐴_𝑟,𝑛. Obviously, when 𝑛=2,𝑟≠0, or 𝑛=1, 𝐴_𝑟,𝑛 is also an invertible matrix.

Theorem 8. Let 𝐴_𝑟,𝑛=Circ(𝐹_𝑟+1,𝐹_𝑟+2,…,𝐹_𝑟+𝑛) be a circulant matrix. Then one has det𝐴_𝑟,𝑛=𝐹_𝑟+1⋅𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1×𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−2, (9) where 𝐹_𝑟+𝑛 is the (𝑟+𝑛)th Fibonacci number. Specially, when 𝑟=0, this result is the same as Theorem 2.1 in [26].

Proof. Obviously, det𝐴₁=(1−𝐹_𝑛+1)^𝑛−1+𝐹_𝑛^𝑛−2∑^𝑛−1_𝑘=1𝐹_𝑘((1−𝐹_𝑛+1)/𝐹_𝑛)^𝑘−1 satisfies the formula. In the case 𝑛>1, let 1−𝐹Γ=^𝑟+2𝐹_𝑟+11,Π−11−1001−1−1⋮⋰⋰⋰01⋰⋰01−1⋰001−1−1₁=0𝐹100⋯00_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−20𝐹0⋯01_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−30𝐹0⋯10⋮⋮⋮⋱⋮⋮_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+11⋯00010⋯00 (10) be two 𝑛×𝑛 matrices; then we have Γ𝐴_𝑟,𝑛Π₁=𝐹_𝑟+1𝑓^′_𝑟,𝑛𝐹_{𝑟+𝑛−1}⋯𝐹_𝑟+20𝑓_𝑟,𝑛𝜏_𝑛⋯𝜏₃00𝐹_𝑟+1−𝐹_𝑟+𝑛+1000𝐹_𝑟−𝐹_𝑟+𝑛0⋮⋮⋱000𝐹_𝑟+1−𝐹_𝑟+𝑛+1, (11) where 𝑎_𝑟=𝐹_𝑟+2𝐹_𝑟+1,𝑓^′_𝑟,𝑛=^𝑛−1_𝑘=1𝐹_𝑟+𝑘+1𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)},𝜏_𝑘=𝐹_𝑟+𝑘−𝑎_𝑟𝐹_{𝑟+𝑘−1}𝑓,𝑘=3,…,𝑛,_𝑟,𝑛=𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1×𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (12) We obtain detΓdet𝐴_𝑟,𝑛detΠ₁=𝐹_𝑟+1⋅𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1×𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−2; (13) while detΓ=(−1)^{(𝑛−1)(𝑛−2)/2},detΠ₁=(−1)^{(𝑛−1)(𝑛−2)/2}, (14) we have det𝐴_𝑟,𝑛=𝐹_𝑟+1⋅𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1×𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−2. (15)

Theorem 9. Let 𝐴_𝑟,𝑛=Circ(𝐹_𝑟+1,𝐹_𝑟+2,…,𝐹_𝑟+𝑛) be a circulant matrix; if 𝑛>2, then 𝐴_𝑟,𝑛 is an invertible matrix. Specially, when 𝑟=0, one gets Theorem 2.2 in [26].

Proof. When 𝑛 = 3 in Theorem 8, then we have det𝐴_𝑟,𝑛=(𝐹_𝑟+1+𝐹_𝑟+2+𝐹_𝑟+3)(𝐹²_𝑟+1+𝐹_𝑟𝐹_𝑟+2)≠0; hence 𝐴_𝑟,𝑛 is invertible. In the case 𝑛>3, since 𝐹_𝑟+𝑛=(𝛼^𝑟+𝑛−𝛽^𝑟+𝑛)/(𝛼−𝛽), where 𝛼+𝛽=1, 𝛼𝛽=−1. We have 𝑓𝜔^𝑘=^𝑛_𝑗=1𝐹_𝑟+𝑗𝜔^𝑘𝑗−1=1𝛼−𝛽^𝑛_𝑗=1𝛼^𝑟+𝑗−𝛽^𝑟+𝑗𝜔^𝑘𝑗−1=1𝛼𝛼−𝛽^𝑟+11−𝛼^𝑛1−𝛼𝜔^𝑘−𝛽^𝑟+11−𝛽^𝑛1−𝛽𝜔^𝑘=𝛼^𝑟+1−𝛽^𝑟+1−𝛼^𝑟+𝑛+1−𝛽^𝑟+𝑛+1𝛼−𝛽1−𝛼+𝛽𝜔^𝑘+𝛼𝛽𝜔^2𝑘−𝛼𝛽𝛼^𝑟−𝛽^𝑟−𝛼^𝑟+𝑛+𝛽^𝑟+𝑛𝜔^𝑘𝛼−𝛽1−𝛼+𝛽𝜔^𝑘+𝛼𝛽𝜔^2𝑘=𝐹_𝑟+1−𝐹_𝑟+𝑛+1+𝐹_𝑟−𝐹_𝑟+𝑛𝜔^𝑘1−𝜔^𝑘−𝜔^2𝑘(𝑘=1,2,…,𝑛−1). (16) If there exists 𝜔^𝑙(𝑙=1,2,…,𝑛−1) such that 𝑓(𝜔^𝑙)=0, we obtain 𝐹_𝑟+1−𝐹_𝑟+𝑛+1+(𝐹_𝑟−𝐹_𝑟+𝑛)𝜔^𝑙=0 for 1−𝜔^𝑙−𝜔^2𝑙≠0; thus, 𝜔^𝑙=(𝐹_𝑟+1−𝐹_𝑟+𝑛+1)/(𝐹_𝑟+𝑛−𝐹_𝑟) is a real number. While 𝜔^𝑙=exp2𝑙𝜋𝑖𝑛=cos2𝑙𝜋𝑛+𝑖sin2𝑙𝜋𝑛, (17) hence, sin(2𝑙𝜋/𝑛)=0; so we have 𝜔^𝑙=−1 for 0<2𝑙𝜋/𝑛<2𝜋. But 𝑥=−1 is not the root of the equation 𝐹_𝑟+1−𝐹_𝑟+𝑛+1+(𝐹_𝑟−𝐹_𝑟+𝑛)𝑥=0(𝑛>3). We obtain 𝑓(𝜔^𝑘)≠0 for any 𝜔^𝑘(𝑘=1,2,…,𝑛−1), while 𝑓(1)=∑^𝑛_𝑗=1𝐹_𝑟+𝑗=−𝐹_𝑟+1+𝐹_𝑟+𝑛+1−(𝐹_𝑟−𝐹_𝑟+𝑛)=𝐹_𝑟+𝑛+2−𝐹_𝑟+2≠0. By Lemma 4, the proof is completed.

Lemma 10. Let the entries of the matrix 𝒢=[𝑔_𝑖,𝑗]^𝑛−2_𝑖,𝑗=1 be of the form 𝑔_𝑖,𝑗=𝐹_𝑟+1−𝐹_𝑟+𝑛+1𝐹,𝑖=𝑗,_𝑟−𝐹_𝑟+𝑛,𝑖=𝑗+1,0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒; (18) then the entries of the inverse 𝒢⁻¹=[𝑔^′_𝑖,𝑗]^𝑛−2_𝑖,𝑗=1 of the matrix 𝒢 are equal to 𝑔^′_𝑖,𝑗=𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−𝑗𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑖−𝑗+1},𝑖≥𝑗,0,𝑖<𝑗. (19) In particular, when 𝑟=0, one gets Lemma 2.1 in [26].

Proof. Let 𝑐_𝑖,𝑗=∑^𝑛−2_𝑘=1𝑔_𝑖,𝑘𝑔^′_𝑘,𝑗. Obviously, 𝑐_𝑖,𝑗=0 for 𝑖<𝑗. In the case 𝑖=𝑗, we obtain 𝑐_𝑖,𝑖=𝑔_𝑖,𝑖𝑔^′_𝑖,𝑖=𝐹_𝑟+1−𝐹_𝑟+𝑛+1⋅1𝐹_𝑟+1−𝐹_𝑟+𝑛+1=1. (20) For 𝑖≥𝑗+1, we obtain 𝑐_𝑖,𝑗=^𝑛−2_𝑘=1𝑔_𝑖,𝑘𝑔^′_𝑘,𝑗=𝑔_{𝑖,𝑖−1}𝑔^′_{𝑖−1,𝑗}+𝑔_𝑖,𝑖𝑔^′_𝑖,𝑗=𝐹_𝑟−𝐹_𝑟+𝑛⋅𝐹_𝑟+𝑛−𝐹_𝑟^{𝑖−𝑗−1}𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖−𝑗+𝐹_𝑟+1−𝐹_𝑟+𝑛+1⋅𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−𝑗𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑖−𝑗+1}=0. (21) Hence, we verify 𝒢𝒢⁻¹=𝐼_𝑛−2, where 𝐼_𝑛−2 is (𝑛−2)×(𝑛−2) identity matrix. Similarly, we can verify 𝒢⁻¹𝒢=𝐼_𝑛−2. Thus, the proof is completed.

Theorem 11. Let 𝐴_𝑟,𝑛=𝐶𝑖𝑟𝑐(𝐹_𝑟+1,𝐹_𝑟+2,…,𝐹_𝑟+𝑛)(𝑛>2) be a circulant matrix.
Then one has 𝐴⁻¹_𝑟,𝑛=1𝑓_𝑟,𝑛×Circ1+^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+2−𝑖}−𝐹_𝑟+2𝐹_𝑟+1𝐹_{𝑟+𝑛+1−𝑖}×𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−1×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖−1𝐹,−^𝑟+2𝐹_𝑟+1+^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+1−𝑖}−𝐹_𝑟+2/𝐹_𝑟+1𝐹_{𝑟+𝑛−𝑖}×𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−1×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖,−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+1−𝐹_𝑟+𝑛+1,−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1²,−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟²𝐹_𝑟+1−𝐹_𝑟+𝑛+1³−𝐹,…,_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟^𝑛−3𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−2, (22) where 𝑓_𝑟,𝑛=𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (23) Specially, when 𝑟=0, this result is the same as Theorem 2.3 in [26].

Proof. Let Π₂=𝑓1−^′𝑟,𝑛𝐹_𝑟+1𝑥₃𝑥₄⋯𝑥_𝑛01𝑦₃𝑦₄⋯𝑦_𝑛,0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1 (24) where 𝑎_𝑟=𝐹_𝑟+2/𝐹_𝑟+1, 𝑥_𝑖=𝑓^′_𝑟,𝑛𝑓_𝑟,𝑛𝐹_{𝑟+𝑛+3−𝑖}−𝐹_𝑟+2/𝐹_𝑟+1𝐹_{𝑟+𝑛+2−𝑖}𝐹_𝑟+1−𝐹_{𝑟+𝑛+2−𝑖}𝐹_𝑟+1(),𝑦𝑖=3,4,…,𝑛_𝑖𝐹=−^{𝑟+𝑛+3−𝑖}−𝐹_𝑟+2/𝐹_𝑟+1𝐹_{𝑟+𝑛+2−𝑖}𝑓_𝑟,𝑛(),𝑓𝑖=3,4,…,𝑛_′𝑟,𝑛=^𝑛−1_𝑘=1𝐹_𝑟+𝑘+1𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)},𝑓_𝑟,𝑛=𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (25) We have Γ𝐴_𝑟,𝑛Π₁Π₂=𝒟₁⊕𝒢, (26) where 𝒟₁=diag(𝐹_𝑟+1,𝑓_𝑟,𝑛) is a diagonal matrix and 𝒟₁⊕𝒢 is the direct sum of 𝒟₁ and 𝒢. If we denote Π=Π₁Π₂, then we obtain 𝐴⁻¹_𝑟,𝑛=Π𝒟₁⁻¹⊕𝒢⁻¹Γ, (27)
and the last row elements of the matrix Π are 0,1,𝑦₃,𝑦₄,…,𝑦_𝑛−1,𝑦_𝑛. By Lemma 10, if let 𝐴⁻¹_𝑟,𝑛=Circ(𝑢₁,𝑢₂,…,𝑢_𝑛), then its last row elements are given by the following equations: 𝑢₂1=−𝑓_𝑟,𝑛𝐹_𝑟+2𝐹_𝑟+1+1𝑓_𝑟,𝑛𝐶_𝑛^(𝑛−2),𝑢₃1=−𝑓_𝑟,𝑛𝐶_𝑛⁽¹⁾,𝑢₄1=−𝑓_𝑟,𝑛𝐶_𝑛⁽²⁾+1𝑓_𝑟,𝑛𝐶_𝑛⁽¹⁾,𝑢₅1=−𝑓_𝑟,𝑛𝐶_𝑛⁽³⁾+1𝑓_𝑟,𝑛𝐶_𝑛⁽²⁾+1𝑓_𝑟,𝑛𝐶_𝑛⁽¹⁾,⋮𝑢_𝑛1=−𝑓_𝑟,𝑛𝐶_𝑛^(𝑛−2)+1𝑓_𝑟,𝑛𝐶_𝑛^(𝑛−3)+1𝑓_𝑛𝐶_𝑛^(𝑛−4),𝑢₁=1𝑓_𝑟,𝑛+1𝑓_𝑟,𝑛𝐶_𝑛^(𝑛−2)+1𝑓_𝑟,𝑛𝐶_𝑛^(𝑛−3). (28)
Let 𝐶_𝑛^(𝑗)=^𝑗_𝑖=1𝐹_{𝑟+3+𝑗−𝑖}−𝐹_𝑟+2/𝐹_𝑟+1𝐹_{𝑟+2+𝑗−𝑖}𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−1𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖=^𝑗_𝑖=1𝑎^′_𝑗,𝑟𝑚_1,𝑟^𝑖𝑚_2,𝑟^𝑖−1𝑗=1,2,…,𝑛−2; (29) we have 𝐶_𝑛⁽²⁾−𝐶_𝑛⁽¹⁾=²_𝑖=1𝑎^′_2,𝑟𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖−𝑎^′_1,𝑟𝑚_1,𝑟=𝑎^′_1,𝑟𝑚_1,𝑟²𝑚_2,𝑟,𝐶_𝑛^(𝑛−2)+𝐶_𝑛^(𝑛−3)=^𝑛−2_𝑖=1𝑎^′_{𝑛−2,𝑟}𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖+^𝑛−3_𝑖=1𝑎^′_{𝑛−3,𝑟}𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖=^𝑛−3_𝑖=1𝐹_{𝑟+𝑛+2−𝑖}−𝐴_𝑟,𝑛𝐹_{𝑟+𝑛+1−𝑖}𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖+𝑎^′_1,𝑟𝑚_1,𝑟^𝑛−2𝑚_2,𝑟^𝑛−3=^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+2−𝑖}−𝐹_𝑟+2/𝐹_𝑟+1𝐹_{𝑟+𝑛+1−𝑖}𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖,𝐶_𝑛^(𝑗+2)−𝐶_𝑛^(𝑗+1)−𝐶_𝑛^(𝑗)=^𝑗+2_𝑖=1𝑎^′_𝑗+2,𝑟𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖−^𝑗+1_𝑖=1𝑎^′_𝑗+1,𝑟𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖−^𝑗_𝑖=1𝑎^′_𝑗,𝑟𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖=𝐹_𝑟+4−𝐴_𝑟,𝑛𝐹_𝑟+3𝑚_2,𝑟^𝑗𝑚_1,𝑟^𝑗+1+𝐹_𝑟+3−𝐴_𝑟,𝑛𝐹_𝑟+2𝑚_2,𝑟^𝑗+1𝑚_1,𝑟^𝑗+2−𝐹_𝑟+3−𝐴_𝑟,𝑛𝐹_𝑟+2𝑚_2,𝑟^𝑗𝑚_1,𝑟^𝑗+1=𝐹_𝑟+3−𝐴_𝑟,𝑛𝐹_𝑟+2𝑚_2,𝑟^𝑗+1𝑚_1,𝑟^𝑗+2𝑗=1,2,…,𝑛−4. (30) We obtain 𝐴⁻¹_𝑟,𝑛=Circ1+𝐶_𝑛^(𝑛−2)+𝐶_𝑛^(𝑛−3)𝑓_𝑟,𝑛,𝐶_𝑛^(𝑛−2)−𝐹_𝑟+2/𝐹_𝑟+1𝑓_𝑟,𝑛,−𝐶_𝑛⁽¹⁾𝑓_𝑟,𝑛𝐶,−^𝑛(2)−𝐶_𝑛⁽¹⁾𝑓_𝑟,𝑛,−𝐶_𝑛⁽³⁾−𝐶_𝑛⁽²⁾−𝐶_𝑛⁽¹⁾𝑓_𝑟,𝑛−𝐶,…,_{𝑛(𝑛−2)}−𝐶_𝑛^(𝑛−3)−𝐶_𝑛^(𝑛−4)𝑓_𝑟,𝑛=1𝑓_𝑟,𝑛Circ1+^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+2−𝑖}−𝐹_𝑟+2𝐹_𝑟+1𝐹_{𝑟+𝑛+1−𝑖}×𝑚_2,𝑟^𝑖−1×𝑚_1,𝑟^𝑖−1,−𝐹_𝑟+2𝐹_𝑟+1+^𝑛−2_𝑖=1𝑎^′_{𝑛−2,𝑟}𝑚_2,𝑟^𝑖−1𝑚_1,𝑟^𝑖,−𝑎^′_1,𝑟𝑚_1,𝑟𝑎,−^′1,𝑟𝑚_1,𝑟²𝑚_2,𝑟,−𝑎^′_1,𝑟𝑚_1,𝑟³𝑚_2,𝑟²−𝑎,…,_′1,𝑟𝑚_1,𝑟^𝑛−2𝑚_2,𝑟^𝑛−3=1𝑓_𝑟,𝑛Circ1+^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+2−𝑖}−𝐹_𝑟+2𝐹_𝑟+1𝐹_{𝑟+𝑛+1−𝑖}×𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−1×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖−1𝐹,−^𝑟+2𝐹_𝑟+1+^𝑛−2_𝑖=1𝐹_{𝑟+𝑛+1−𝑖}−𝐹_𝑟+2𝐹_𝑟+1𝐹_{𝑟+𝑛−𝑖}×𝐹_𝑟+𝑛−𝐹_𝑟^𝑖−1×𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑖−1−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+1−𝐹_𝑟+𝑛+1,−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1²,−𝐹_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟²𝐹_𝑟+1−𝐹_𝑟+𝑛+1³−𝐹,…,_𝑟+3−𝐹_𝑟+2/𝐹_𝑟+1𝐹_𝑟+2𝐹_𝑟+𝑛−𝐹_𝑟^𝑛−3𝐹_𝑟+1−𝐹_𝑟+𝑛+1^𝑛−2, (31) where 𝑓_𝑟,𝑛=𝐹_𝑟+1−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑛+^𝑛−2_𝑘=1𝐹_𝑟+𝑘+2−𝐹_𝑟+2𝐹_𝑟+1𝐹_𝑟+𝑘+1𝐹_𝑟+𝑛−𝐹_𝑟𝐹_𝑟+1−𝐹_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (32)

3. Determinant, Invertibility, and Inverse of Circulant Matrix with Any Continuous Lucas Numbers

In this section, let 𝐵_𝑟,𝑛=Circ(𝐿_𝑟+1,𝐿_𝑟+2,…,𝐿_𝑟+𝑛) be a circulant matrix. Firstly, we give a determinant formula for the matrix 𝐵_𝑟,𝑛. Afterwards, we prove that 𝐵_𝑟,𝑛 is an invertible matrix for any positive integer 𝑛, and then we find the inverse of the matrix 𝐵_𝑟,𝑛.

Theorem 12. Let 𝐵_𝑟,𝑛=Circ(𝐿_𝑟+1,𝐿_𝑟+2,…,𝐿_𝑟+𝑛) be a circulant matrix; then one has det𝐵_𝑟,𝑛=𝐿_𝑟+1⋅𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑘+1×𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−2, (33) where 𝐿_𝑟+𝑛 is the (𝑟+𝑛)th Lucas number. In particular, when 𝑟=0, one gets Theorem 3.1 in [26].

Proof. Obviously, det𝐵₁=(1−𝐿_𝑛+1)^𝑛−1+(𝐿_𝑛−2)^𝑛−2∑^𝑛−1_𝑘=1(𝐿_𝑘+2−3𝐿_𝑘+1)((1−𝐿_𝑛+1)/(𝐿_𝑛−2))^𝑘−1 satisfies the formula, when 𝑛>1; let 1−𝐿Σ=^𝑟+2𝐿_𝑟+11,Ω−11−1001−1−1⋮⋰⋰⋰01⋰⋰01−1⋰001−1−1₁=0𝐿100⋯00_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−20𝐿0⋯01_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−30𝐿0⋯10⋮⋮⋮⋱⋮⋮_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1,1⋯00010⋯00 (34) be two 𝑛×𝑛 matrices, we have Σ𝐵_𝑟,𝑛Ω₁=𝐿_𝑟+1𝑙^′_𝑟,𝑛𝐿_{𝑟+𝑛−1}⋯𝐿_𝑟+20𝑙_𝑟,𝑛𝐿_𝑟+𝑛−𝑏_𝑟𝐿_{𝑟+𝑛−1}⋯𝐿_𝑟+3−𝑏_𝑟𝐿_𝑟+200𝐿_𝑟+1−𝐿_{𝑟+𝑛−1}000𝐿_𝑟−𝐿_𝑟+𝑛0⋮⋮⋱000𝐿_𝑟+1−𝐿_{𝑟+𝑛−1}, (35) where 𝑏_𝑟=𝐿_𝑟+2𝐿_𝑟+1,𝑙_𝑟,𝑛=𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑘+1×𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)},𝑙^′_𝑟,𝑛=^𝑛−1_𝑘=1𝐿_𝑟+𝑘+1𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (36) We obtain detΣdet𝐵_𝑟,𝑛detΩ₁=𝐿_𝑟+1⋅𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝑏_𝑟𝐿_𝑟+𝑘+1×𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−2, (37) while detΣ=detΩ₁=(−1)^{(𝑛−1)(𝑛−2)/2}. (38) We have det𝐵_𝑟,𝑛=𝐿_𝑟+1⋅𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑘+1×𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−2. (39)

Theorem 13. Let 𝐵_𝑟,𝑛=Circ(𝐿_𝑟+1,𝐿_𝑟+2,…,𝐿_𝑟+𝑛) be a circulant matrix; then 𝐵_𝑟,𝑛 is invertible for any positive integer 𝑛. Specially, when 𝑟=0, one gets Theorem 3.2 in [26].

Proof. Since 𝐿_𝑛+𝑟=𝛼^𝑛+𝑟+𝛽^𝑛+𝑟, where 𝛼+𝛽=1, 𝛼⋅𝛽=−1. Hence we have 𝑓𝜔^𝑘=^𝑛_𝑗=1𝐿_𝑛+𝑟𝜔^𝑘𝑗−1=^𝑛_𝑗=1𝛼^𝑟+𝑗+𝛽^𝑟+𝑗𝜔^𝑘𝑗−1=𝛼^𝑟+11−𝛼^𝑛1−𝛼𝜔^𝑘+𝛽^𝑟+11−𝛽^𝑛1−𝛽𝜔^𝑘because1−𝛼𝜔^𝑘≠0and1−𝛽𝜔^𝑘=𝛼≠0_𝑟+1+𝛽^𝑟+1−𝛼^𝑟+𝑛+1+𝛽^𝑟+𝑛+11−𝛼+𝛽𝜔^𝑘+𝛼𝛽𝜔^2𝑘−𝛼𝛽𝛼^𝑟+𝛽^𝑟−𝛼^𝑟+𝑛−𝛽^𝑟+𝑛𝜔^𝑘1−𝛼+𝛽𝜔^𝑘+𝛼𝛽𝜔^2𝑘=𝐿_𝑟+1−𝐿_𝑟+𝑛+1+𝐿_𝑟−𝐿_𝑟+𝑛𝜔^𝑘1−𝜔^𝑘−𝜔^2𝑘(𝑘=1,2,…,𝑛−1). (40) If there exists 𝜔^𝑙(𝑙=1,2,…,𝑛−1) such that 𝑓(𝜔^𝑙)=0, we obtain 𝐿_𝑟+1−𝐿_𝑟+𝑛+1+(𝐿_𝑟−𝐿_𝑟+𝑛)𝜔^𝑙=0 for 1−𝜔^𝑙−𝜔^2𝑙≠0; thus, 𝜔^𝑙=(𝐿_𝑟+1−𝐿_𝑟+𝑛+1)/(𝐿_𝑟+𝑛−𝐿_𝑟) is a real number, while 𝜔^𝑙=exp2𝑙𝜋𝑖𝑛=cos2𝑙𝜋𝑛+𝑖sin2𝑙𝜋𝑛. (41)
Hence, sin(2𝑙𝜋/𝑛)=0; we have 𝜔^𝑙=−1 for 0<2𝑙𝜋/𝑛<2𝜋. But 𝑥=−1 is not the root of the equation 𝐿_𝑟+1−𝐿_𝑟+𝑛+1+(𝐿_𝑟−𝐿_𝑟+𝑛)𝑥=0 for any positive integer 𝑛. We obtain 𝑓(𝜔^𝑘)≠0 for any 𝜔^𝑘(𝑘=1,2,…,𝑛−1), while 𝑓(1)=∑^𝑛_𝑗=1𝐿_𝑟+𝑗=𝐿_𝑟+𝑛+2−𝐿_𝑟+2≠0. By Lemma 4, the proof is completed.

Lemma 14. Let the entries of the matrix ℋ=[ℎ_𝑖,𝑗]^𝑛−2_𝑖,𝑗=1 be of the form ℎ_𝑖,𝑗=𝐿_𝑟+1−𝐿_𝑟+𝑛+1𝐿,𝑖=𝑗,_𝑟−𝐿_𝑟+𝑛,𝑖=𝑗+1,0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒; (42) then the entries of the inverse ℋ⁻¹=[ℎ^′_𝑖,𝑗]^𝑛−2_𝑖,𝑗=1 of the matrix ℋ are equal to ℎ^′_𝑖,𝑗=𝐿_𝑟+𝑛−𝐿_𝑟^𝑖−𝑗𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑖−𝑗+1},𝑖≥𝑗,0,𝑖<𝑗. (43) Specially, when 𝑟=0, one gets Lemma 3.1 in [26].

Proof. Let 𝑟_𝑖,𝑗=∑^𝑛−2_𝑘=1ℎ_𝑖,𝑘ℎ^′_𝑘,𝑗. Obviously, 𝑟_𝑖,𝑗=0 for 𝑖<𝑗. In the case 𝑖=𝑗, we obtain 𝑟_𝑖,𝑖=ℎ_𝑖,𝑖ℎ^′_𝑖,𝑖=𝐿_𝑟+1−𝐿_𝑟+𝑛+1⋅1𝐿_𝑟+1−𝐿_𝑟+𝑛+1=1. (44) For 𝑖≥𝑗+1, we obtain 𝑟_𝑖,𝑗=^𝑛−2_𝑘=1ℎ_𝑖,𝑘ℎ^′_𝑘,𝑗=ℎ_{𝑖,𝑖−1}ℎ^′_{𝑖−1,𝑗}+ℎ_𝑖,𝑖ℎ^′_𝑖,𝑗=𝐿_𝑟−𝐿_𝑟+𝑛⋅𝐿_𝑟+𝑛−𝐿_𝑟^{𝑖−𝑗−1}𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑖−𝑗+𝐿_𝑟+1−𝐿_𝑟+𝑛+1⋅𝐿_𝑟+𝑛−𝐿_𝑟^𝑖−𝑗𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑖−𝑗+1}=0. (45) Hence, we verify ℋℋ⁻¹=𝐼_𝑛−2, where 𝐼_𝑛−2 is (𝑛−2)×(𝑛−2) identity matrix. Similarly, we can verify ℋ⁻¹ℋ=𝐼_𝑛−2. Thus, the proof is completed.

Theorem 15. Let 𝐵_𝑟,𝑛=Circ(𝐿_𝑟+1,𝐿_𝑟+2,…,𝐿_𝑟+𝑛) be a circulant matrix; then we have 𝐵⁻¹_𝑟,𝑛=1𝑙_𝑟,𝑛Circ1+^𝑛−2_𝑖=1𝐿_{𝑟+𝑛+2−𝑖}−𝐿_𝑟+2𝐿_𝑟+1𝐿_{𝑟+𝑛+1−𝑖}×𝐿_𝑟+𝑛−𝐿_𝑟^𝑖−1×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑖−1𝐿,−^𝑟+2𝐿_𝑟+1+^𝑛−2_𝑖=1𝐿_{𝑟+𝑛+1−𝑖}−𝐿_𝑟+2𝐿_𝑟+1𝐿_{𝑟+𝑛−𝑖}×𝐿_𝑟+𝑛−𝐿_𝑟^𝑖−1×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑖−1,−𝐿_𝑟+3−𝐿_𝑟+2/𝐿_𝑟+1𝐿_𝑟+2𝐿_𝑟+1−𝐿_𝑟+𝑛+1,−𝐿_𝑟+3−𝐿_𝑟+2/𝐿_𝑟+1𝐿_𝑟+2𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1²,−𝐿_𝑟+3−𝐿_𝑟+2/𝐿_𝑟+1𝐿_𝑟+2𝐿_𝑟+𝑛−𝐿_𝑟²𝐿_𝑟+1−𝐿_𝑟+𝑛+1³−𝐿,…,_𝑟+3−𝐿_𝑟+2/𝐿_𝑟+1𝐿_𝑟+2𝐿_𝑟+𝑛−𝐿_𝑟^𝑛−3𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑛−2, (46) where 𝑙_𝑟,𝑛=𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑘+1𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (47) In particular, when 𝑟=0, the result is the same as Theorem 3.3 in [26].

Proof. Let Ω₂ be the form of 𝑙1−^𝑛′𝐿_𝑟+1𝑥^′₃𝑥^′₄⋯𝑥^′_𝑛01𝑦^′₃𝑦^′₄⋯𝑦^′_𝑛,0010⋯00001⋯0⋮⋮⋮⋮⋱⋮0000⋯1 (48) where 𝑏_𝑟=𝐿_𝑟+2𝐿_𝑟+1,𝑥^′_𝑖=𝑙^′_𝑟,𝑛𝑙_𝑟,𝑛𝐿_{𝑟+𝑛+3−𝑖}−𝐿_𝑟+2/𝐿_𝑟+1𝐿_{𝑟+𝑛+2−𝑖}𝐿_𝑟+1−𝐿_{𝑟+𝑛+2−𝑖}𝐿_𝑟+1(),𝑦𝑖=3,4,…,𝑛_′𝑖𝐿=−^{𝑟+𝑛+3−𝑖}−𝐿_𝑟+2/𝐿_𝑟+1𝐿_{𝑟+𝑛+2−𝑖}𝑙_𝑟,𝑛(),𝑙𝑖=3,4,…,𝑛_′𝑟,𝑛=^𝑛−1_𝑘=1𝐿_𝑟+𝑘+1𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)},𝑙_𝑟,𝑛=𝐿_𝑟+1−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑛+^𝑛−2_𝑘=1𝐿_𝑟+𝑘+2−𝐿_𝑟+2𝐿_𝑟+1𝐿_𝑟+𝑘+1𝐿_𝑟+𝑛−𝐿_𝑟𝐿_𝑟+1−𝐿_𝑟+𝑛+1^{𝑛−(𝑘+1)}. (49) We have Σ𝐵_𝑟,𝑛Ω₁Ω₂=𝒟₂⊕ℋ, (50) where 𝒟₂=diag(𝐿_𝑟+1,𝑙_𝑟,𝑛) is a diagonal matrix and 𝒟₂⊕ℋ is the direct sum of 𝒟₂ and ℋ. If we denote Ω=Ω₁Ω₂, we obtain 𝐵⁻¹_𝑟,𝑛=Ω𝒟₂⁻¹⊕ℋ⁻¹Σ, (51)
and the last row elements of the matrix Ω are 0,1,𝑦^′₃,𝑦^′₄,⋯,𝑦^′_𝑛. By Lemma 14, if let 𝐵⁻¹_𝑟,𝑛=Circ(𝑣₁,𝑣₂,…,𝑣_𝑛), then its last row elements are given by the following equations: 𝑣₂1=−𝑙_𝑟,𝑛𝐿_𝑟+2𝐿_𝑟+1+1𝑙_𝑟,𝑛𝐷_𝑛^(𝑛−2),𝑣₃1=−𝑙_𝑟,𝑛𝐷_𝑛⁽¹⁾,𝑣₄1=−𝑙_𝑟,𝑛𝐷_𝑛⁽²⁾+1𝑙_𝑟,𝑛𝐶_𝑛⁽¹⁾,𝑣₅1=−𝑙_𝑟,𝑛𝐷_𝑛⁽³⁾+1𝑙_𝑟,𝑛𝐷_𝑛⁽²⁾+1𝑙_𝑟,𝑛𝐷_𝑛⁽¹⁾,⋮𝑣_𝑛1=−𝑙_𝑟,𝑛𝐷_𝑛^(𝑛−2)+1𝑙_𝑟,𝑛𝐷_𝑛^(𝑛−3)+1𝑙_𝑛𝐷_𝑛^(𝑛−4),𝑣₁=1𝑙_𝑟,𝑛+1𝑙_𝑟,𝑛𝐷_𝑛^(𝑛−2)+1𝑙_𝑟,𝑛𝐷_𝑛^(𝑛−3). (52)
Let 𝐷_𝑛^(𝑗)=^𝑗_𝑖=1𝐿_{𝑟+3+𝑗−𝑖}−𝐿_𝑟+2𝐿_𝑟+1𝐿_{𝑟+2+𝑗−𝑖}×𝐿_𝑟+𝑛−𝐿_𝑟^𝑖−1×𝐿_𝑟+1−𝐿_𝑟+𝑛+1^𝑖−1=^𝑗_𝑖=1𝑏^′_𝑗,𝑟ℎ_1,𝑟^𝑖ℎ_2,𝑟^𝑖−1𝑗=1,2,…,𝑛−2; (53) we have 𝐷_𝑛⁽²⁾−𝐷_𝑛⁽¹⁾=²_𝑖=1𝑏^′_2,𝑟ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖−𝑏^′_1,𝑟ℎ_1,𝑟=𝑏^′_1,𝑟ℎ_1,𝑟²ℎ_2,𝑟,𝐷_𝑛^(𝑛−2)+𝐷_𝑛^(𝑛−3)=^𝑛−2_𝑖=1𝑏^′_{𝑛−2,𝑟}ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖+^𝑛−3_𝑖=1𝑏^′_{𝑛−3,𝑟}ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖=^𝑛−3_𝑖=1𝐿_{𝑟+𝑛+2−𝑖}−𝐵_𝑟,𝑛𝐿_{𝑟+𝑛+1−𝑖}ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖+𝑏^′_1,𝑟ℎ_1,𝑟^𝑛−2ℎ_2,𝑟^𝑛−3=^𝑛−2_𝑖=1𝐿_{𝑟+𝑛+2−𝑖}−𝐵_𝑟,𝑛𝐿_{𝑟+𝑛+1−𝑖}ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖,𝐷_𝑛^(𝑗+2)−𝐷_𝑛^(𝑗+1)−𝐷_𝑛^(𝑗)=^𝑗+2_𝑖=1𝑏^′_𝑗+2,𝑟ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖−^𝑗+1_𝑖=1𝑏^′_𝑗+1,𝑟ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖−^𝑗_𝑖=1𝑏^′_𝑗,𝑟ℎ_2,𝑟^𝑖−1ℎ_1,𝑟^𝑖=𝐿_𝑟+4−𝐵_𝑟,𝑛𝐿_𝑟+3ℎ_2,𝑟^𝑗ℎ_1,𝑟^𝑗+1+𝐿_𝑟+3−𝐵_𝑟,𝑛𝐿_𝑟+2ℎ_2,𝑟^𝑗+1ℎ_1,𝑟^𝑗+2−𝐿_𝑟+3−𝐵_𝑟,𝑛𝐿_𝑟+2ℎ_2,𝑟^𝑗ℎ_1,𝑟^𝑗+1=𝐿_𝑟+3−𝐵_𝑟,𝑛𝐿_𝑟+2ℎ_2,𝑟^𝑗+1ℎ_1,𝑟^𝑗+2=𝑏^′_1,𝑟ℎ_2,𝑟^𝑗+1ℎ_1,𝑟^𝑗+2𝑗=1,2,…,𝑛−4. (54) We obtain 𝐵⁻¹_𝑟,𝑛=Circ1+𝐷_𝑛^(𝑛−3)+𝐷_𝑛^(𝑛−2)𝑙_𝑟,𝑛,𝐷_𝑛^(𝑛−2)−𝐿_𝑟+2/𝐿_𝑟+1𝑙_𝑟,𝑛,