# Explicit Form of the Inverse Matrices of Tribonacci Circulant Type Matrices

Authors: Li Liu; Zhaolin Jiang
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

It is a hot topic that circulant type matrices are applied to networks engineering. The determinants
and inverses of Tribonacci circulant type matrices are discussed in the paper. Firstly, Tribonacci
circulant type matrices are defined. In addition, we show the invertibility of Tribonacci circulant
matrix and present the determinant and the inverse matrix based on constructing the transformation
matrices. By utilizing the relation between left circulant,g-circulant matrices and circulant matrix,
the invertibility of Tribonacci left circulant and Tribonacci g-circulant matrices is also discussed.
Finally, the determinants and inverse matrices of these matrices are given, respectively.


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## Body

## 1. Introduction

Circulant type matrices have important applications in various networks engineering. Exploiting the circulant structure of the channel matrices, Eghbali et al. [1] analysed the realistic near fast fading scenarios with circulant frequency selective channels. The optimum sampling in the one- and two-dimensional (1D and 2D) wireless sensor networks (WSNs) with spatial temporally correlated data was studied with circulant matrices in [2]. The repeat space theory (RST) was extended to apply to carbon nanotubes and related molecular networks, where the corresponding matrices are pseudocirculant in [3]. Preconditioners obtained by circulant approximations of stochastic automata networks were considered in [4]. In [5], circulant mutation whose differential equations obtained neither are of repliator-type nor can they be transformed straightway into a linear equation was introduced into autocatalytic reaction networks. Jing and Jafarkhani [6] proposed distributed differential space-time codes that work for networks with any number of relays using circulant matrices. Wang and Cheng [7] studied the existence of doubly periodic travelling waves in cellular networks involving the discontinuous Heaviside step function by circulant matrix. Pais et al. [8] proved conditions for the existence of stable limit cycles arising from multiple distinct Hopf bifurcations of the dynamics in the case of circulant fitness matrices.Circulant type matrices have been put on the firm basis with the work in [9, 10] and so on. Furthermore, the g-circulant matrices are focused on by many researchers; for the details please refer to [11–13] and the references therein.Lately, some scholars gave the explicit determinant and inverse of the circulant and skew-circulant involving famous numbers. Jiang et al. [14] discussed the invertibility of circulant type matrices with the sum and product of Fibonacci and Lucas numbers and presented the determinants and the inverses of these matrices. Jiang et al. [15] considered circulant type matrices with the k-Fibonacci and k-Lucas numbers and presented the explicit determinant and inverse matrix by constructing the transformation matrices. Jiang and Hong [16] gave exact form determinants of the RSFPLR circulant matrices and the RSLPFL circulant matrices involving Perrin, Padovan, Tribonacci, and the generalized Lucas number by the inverse factorization of polynomial. Bozkurt and Tam gave determinants and inverses of circulant matrices with Jacobsthal and Jacobsthal-Lucas numbers in [17]. Cambini presented an explicit form of the inverse of a particular circulant matrix in [18]. Shen et al. considered circulant matrices with Fibonacci and Lucas numbers and presented their explicit determinants and inverses in [19].The Tribonacci sequences are defined by the following recurrence relations [20–22], respectively: (1)Tn=Tn-1+Tn-2+Tn-3,where T0=0, T1=1, T2=1, n≥3.The first few values of the sequences are given by the following table:(2)n012345678⋯Tn011247132444⋯.Let t1, t2, and t3 be the roots of the characteristic equation x3-x2-x-1=0 and then we have (3)t1+t2+t3=1,t1t2+t1t3+t2t3=-1,t1t2t3=1.Hence the Binet formulas of the sequences {Tn} have the form (4)Tn=b1t1n+b2t2n+b3t3n,where bi is the ith root of the polynomial 44y3-2y-1 for i=1,2,3.In this paper, we consider circulant type matrices, including the circulant and left circulant andg-circulant matrices. If we suppose Tn is the nth Tribonacci number, then we define a Tribonacci circulant matrix which is an n×n matrix with the following form: (5)CircT1,T2,⋯,Tn=T1T2⋯TnTnT1⋯Tn-1⋮⋮⋮T2T3⋯T1.Besides, a Tribonacci left circulant matrix is given by(6)LCircT1,T2,⋯,Tn=T1T2⋯TnT2T3⋯T1⋮⋮⋮TnT1⋯Tn-1,where each row is a cyclic shift of the row above to the left.A Tribonaccig-circulant matrix is an n×n matrix with the following form: (7)Ag,n=T1T2⋯TnTn-g+1Tn-g+2⋯Tn-gTn-2g+1Tn-2g+2⋯Tn-2g⋮⋮⋱⋮Tg+1Tg+2⋯Tg,where g is a nonnegative integer and each of the subscripts is understood to be reduced modulo n. The first row of Ag,n is T1,T2,…,Tn, and its j+1th 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 standard Tribonacci circulant matrix. If g=n-1, then we obtain Tribonacci left circulant matrix.Lemma 1.
Then×n tridiagonal matrix is given by(8)An=τ2τ10τ3τ2τ1τ3τ2τ1⋱⋱⋱τ3τ2τ1τ3τ2τ10τ3τ2;
then(9)det⁡An=τ2+τ22-4τ1τ3/2n+1-τ2-τ22-4τ1τ3/2n+1τ22-4τ1τ3,τ22≠4τ1τ3τ22≠4τ1τ3,τ22≠4τ1τ3,τ22≠4τ1τ300,τ22≠4τ1τ3,n+1τ22n,τ22=4τ1τ3.Proof.
det
⁡
A
n
=
τ
2
·
det
⁡
A
n
-
1
-
τ
1
τ
3
·
det
⁡
A
n
-
2; let x+y=τ2, xy=τ1τ3 and then let x, y be the roots of the equation x2-τ2x+τ1τ3=0.
We have(10)det⁡An=yn+xyn-1+⋯+xn-1y+xn=xn+1-yn+1x-y,x≠y,n+1xn,x=y,where x=τ2+τ22-4τ1τ3/2 and y=τ2-τ22-4τ1τ3/2.
Hence,(11)det⁡An=τ2+τ22-4τ1τ3/2n+1-τ2-τ22-4τ1τ3/2n+1τ22-4τ1τ3,,τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ3.0000τ22≠4τ1τ3,n+1τ22n,τ22=4τ1τ3.Lemma 2.
Let(12)Bn=a1a2a3⋯an-2an-1anτ2τ1τ3τ2τ1τ3τ2τ1⋱⋱⋱τ3τ2τ1τ3τ2τ1be an n×n matrix; one has (13)det⁡Bn=∑i=1n‍-11+iτ1n-iai·det⁡Ai-1,where (14)det⁡Ai-1=τ2+τ22-4τ1τ3/2i-τ2-τ22-4τ1τ3/2iτ22-4τ1τ3,,τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ3.τ22≠4τ1τ3,iτ22i-1,τ22=4τ1τ3.Specifically, det⁡A0=1.Proof.
Accoding to the last column determinant expansion and Lemma1, we obtain(15)det⁡Bn=τ1·det⁡Bn-1+-1n+1an·det⁡An-1=τ1τ1·det⁡Bn-2+-1nan-1·det⁡An-2+-1n+1an·det⁡An-1=τ1n-1·det⁡B1+-11+2τ1n-2a2·det⁡A1+-11+3τ1n-3a3·det⁡A2+⋯+-11+nan·det⁡An-1=∑i=1n‍-11+iτ1n-iai·det⁡Ai-1.


## 2. Determinant and Inverse of Tibonacci Circulant Matrix

In this section, letDn=Circ(T1,T2,…,Tn) be a Tribonacci circulant matrix. Firstly, we give the determinant of the matrix Dn. Afterwards, we discuss the invertibility of the matrix Dn, and we find the inverse of the matrix Dn.Theorem 3.
LetDn=Circ(T1,T2,…,Tn) be a Tribonacci circulant matrix; then we have (16)det⁡Dn=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-iδ1--Tn+ΔT1-Tn-Tn-1+∑i=1n-3Δi+1Tn-i-2‍δ2,where Tn is the nth Tribonacci number, and(17)Δ=-b±b2-4ac2a,δ1=T1-Tn-Tn-1T1-Tn+1n-3+∑i=2n-2‍-11+iT1-Tn+1n-i-2Tn-i-1·det⁡Ai-1,δ2=∑i=1n-2‍-11+iT1-Tn+1n-i-2Tn-i+1-Tn-i·det⁡Ai-1,det⁡Ai-1=b+b2-4ac/2i-b-b2-4ac/2ib2-4ac,τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ34τ1τ3.b2≠4ac,ib2i-1,b2=4ac,a=T1-Tn+1,b=-Tn-Tn-1,c=-Tn.Proof.
Obviously,det⁡D1=1 satisfies (16). In the case where n>1, let(18)Γ1=10-11-11-1-11-1-101-1-1-1⋮⋰⋰⋰⋰01-1-1-101-1-1-1,Π1=100⋯000Δn-20⋯010Δn-30⋯10⋮⋮⋮⋰⋮⋮0Δ1⋯00010⋯00be two n×n matrices and then we have(19)Γ1DnΠ1=T1p1Tn-1Tn-2⋯T3T20p2φ3φ4⋯φn-1φn0p3ϕTn-3⋯T2T100ba00cba⋱⋱⋱cba,where (20)p1=∑i=1n-1‍Ti+1Δn-i+1,p2=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-i,p3=-Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2,φi=Tn+3-i-Tn+2-i,i=3,…,n,ϕ=T1-Tn-Tn-1,Δ=-b±b2-4ac2a,a=T1-Tn+1,b=-Tn-Tn-1,c=-Tn.We obtain (21)det⁡Γ1det⁡Dndet⁡Π1=T1·p2δ1-p3δ2=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-iδ1--Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2δ2,where(22)δ1=T1-Tn-Tn-1T1-Tn+1n-3+∑i=2n-2‍-11+iT1-Tn+1n-i-2·Tn-i-1·det⁡Ai-1,δ2=∑i=1n-2‍-11+iT1-Tn+1n-i-2Tn-i+1-Tn-i·det⁡Ai-1.Let (23)Bn=ϕTn-3⋯T3T2T1bacba⋱⋱⋱cbacba,Cn=φ3φ4⋯φn-2φn-1φnbacba⋱⋱⋱cbacbabe two (n-2)×(n-2) matrices, and δ1=det⁡Bn, and δ2=det⁡Cn.
Aoccording to Lemma2, thus(24)δ1=T1-Tn-Tn-1T1-Tn+1n-3+∑i=2n-2‍-11+iT1-Tn+1n-i-2·Tn-i-1·det⁡Ai-1,δ2=∑i=1n-2‍-11+iT1-Tn+1n-i-2Tn-i+1-Tn-i·det⁡Ai-1.While (25)det⁡Γ1=det⁡Π1=-1n-1n-2/2,we have (26)det⁡Dn=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-iδ1--Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2δ2.Theorem 4.
LetDn=Circ(T1,T2,…,Tn) be a Tribonacci circulant matrix; if n≠2 and n≠2kπarctan⁡±4ac-b2/-b-1(k=1,2,…,n-1), then Dn is an invertible matrix.Proof.
Whenn=2 in Theorem 3, then we have det⁡D2=0. Hence, D2 is not invertible.
In the case wheren>2, since Tn=b1t1n+b2t2n+b3t3n, where bi is the ith root of the polynomial 44y3-2y-1, we have (27)fωk=∑j=1n‍Tjωkj-1=∑j=1n‍b1t1j+b2t2j+b3t3jωkj-1=b1t11-t1n1-t1ωk+b2t21-t2n1-t2ωk+b3t31-t3n1-t3ωk=1-Tn+1+-Tn-Tn-1ωk-Tnω2k1-ωk-ω2k-ω3kvvvvvvvvvvvvvvvk=1,2,…,n-1.If there exists ωl(l=1,2,…,n-1) such that f(ωl)=0, we obtain 1-Tn+1+(-Tn-Tn-1)ωk-Tnω2k=0 for 1-ωk-ω2k-ω3k≠0. Thus, ωl=(Tn+Tn-1±(-Tn-Tn-1)2+4Tn(T1-Tn+1))/-2Tn.
Leta=T1-Tn+1, b=-Tn-Tn-1, and c=-Tn; if b2-4ac≥0, we have ωl is a real number.
While(28)ωl=exp⁡2lπin=cos⁡2lπn+isin2lπn,sin(2lπ/n)=0. We have ωl=-1 for 0<(2lπ/n)<2π, but 1-ωk-ω2k-ω3k=0. We obtain f(ωk)≠0 for any ωk(k=1,2,…,n-1), while f(1)=∑j=1n‍Tj=-(1/2)(1-Tn-Tn+2)≠0.
Ifb2-4ac<0, ωl is an imaginary number.
If(29)cos⁡2lπn=Tn+Tn-1-2Tnsin2lπn=-4TnT1-Tn+1--Tn-Tn-12-2Tnor (30)cos⁡2lπn=Tn+Tn-1-2Tn,sin2lπn=--4TnT1-Tn+1--Tn-Tn-12-2Tn,we obtain n=2kπ(arctan(±4ac-b2/-b))-1, such that f(ωl)=0. If 1-ωk-ω2k-ω3k=0, we have ωk=-1 and if n is an even number, then f(ωk)=∑j=1n‍Tj(ωk)j-1=T1-T2+⋯-Tn<0. By Lemma 1 in [15], the proof is completed.Lemma 5.
LetΦ=aVUA be an n×n matrix; then (31)Φ-1=1l-1lVA-1-1lA-1UA-1+1lA-1UVA-1,where l=a-VA-1U, V is a row vector, and U is a column vector.Proof.
Consider(32)aVUA·1l-1lVA-1-1lA-1UA-1+1lA-1UVA-1=100In-1=In,1l-1lVA-1-1lA-1UA-1+1lA-1UVA-1·aVUA=100In-1=In.Lemma 6.
Let the matrixG=[gi,j]i,j=1n-3 be of the form (33)gi,j=T1-Tn+1,i=j,-Tn-Tn-1,i=j+1,-Tn,i=j+2,0,otherwise;then the inverse G-1=[gi,j′]i,j=1n-3 of the matrix G is equal to (34)gi,j′=1T1-Tn+1βi-j+1-αi-j+1β-α,i≥j,0,i<j,where(35)α=--Tn-Tn-1+-Tn-Tn-12+4TnT1-Tn+12T1-Tn+1,β=--Tn-Tn-1--Tn-Tn-12+4TnT1-Tn+12T1-Tn+1.Proof.
Letci,j=∑k=1n-3‍gi,kgk,j′. Obviously, ci,j=0 for i<j. In the case where i=j, we obtain (36)ci,i=gi,igi,i′=T1-Tn+1·1T1-Tn+1=1.For i≥j+1, we obtain (37)ci,j=∑k=1n-3‍gi,kgk,j′=gi,i-2gi-2,j′+gi,i-1gi-1,j′+gi,igi,j′=-Tn·1T1-Tn+1βi-j-1-αi-j-1β-α+-Tn-Tn-1·1T1-Tn+1βi-j-αi-jβ-α+T1-Tn+1·1T1-Tn+1βi-j+1-αi-j+1β-α=0.Hence, we verify GG-1=In-3, where In-3 is (n-3)×(n-3) identity matrix. Similarly, we can verify G-1G=In-3. Thus, the proof is completed.Theorem 7.
LetDn=Circ(T1,T2,…,Tn) be an invertible Tribonacci circulant matrix; then one has (38)Dn-1=Circx2′+-1-p3p2x3′-x4′-x5′,-x2′+-1+p3p2x3′-x4′,xn′,xn-1′-xn′,xn-2′-xn-1′-xn′,…,x3′-x4′-x5′-x6′-1-p3p2,where (39)p2=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-i,p3=-Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2,x1′=0,x2′=1p2,x3′=γ3l+T1-Tn+1∑i=2n-2gi1γi+2l,x4′=-γ3∑i=2n-3gi1ρi+3l+∑i=1n-3‍gi1γi+3-T1-Tn+1∑i=1n-3gi1ρi+3∑i=2n-2gi1γi+2l,⋮xk′=-γ3∑i=2n-k+1gi1ρi+k-1l+∑i=1n-k+1‍gi1γi+k-1-T1-Tn+1∑i=1n-k+1gi1ρi+k-1∑i=2n-2gi1γi+2l,(k≥4).Proof.
Let(40)Γ2=100⋯0010-p3p2100⋯⋯⋯⋱⋯000⋯1;thus (41)Γ2Γ1DnΠ1=T1p1Tn-1Tn-2⋯T3T20p2φ3φ4⋯φn-1φn00ρ3ρ4⋯ρn-1ρn00ba00cba⋱⋱⋱cba,where (42)ρ3=T1-Tn-Tn-1-p3p2Tn-Tn-1,ρi=Tn-i+1-p3p2Tn-i+3-Tn-i+2i=4,5,⋯,n,φi=Tn+3-i-Tn+2-i,i=3,…,n.According to Lemma 5, let (43)F=ρ3VUGbe an (n-2)×(n-2) matrix and we obtain (44)F-1=1l-VG-1l-G-1UlG-1+UVG-1l,where (45)ρ3=T1-Tn-Tn-1-p3p2Tn-Tn-1,U=-Tn-Tn-1,-Tn,0,…,0T,V=ρ4,ρ5,…,ρn,l=ρ3-b∑i=1n-3‍gi1ρi+3-c∑i=1n-4‍ρi+4.Let (46)Π2=1-p1-Tn-1+p1φ3p2⋯-T2+p1φnp201-φ3p2⋯-φnp2001⋯0⋮⋮⋮⋱⋮000⋯1,where(47)p1=∑i=1n-1‍Ti+1Δn-(i+1),p2=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-i,φi=Tn+3-i-Tn+2-i,i=3,…,n,Δ=-b±b2-4ac2a,a=T1-Tn+1,b=-Tn-Tn-1,c=-Tn.We have (48)Γ2Γ1DnΠ1Π2=D1⊕F,where D1=diag(T1,p2) is a diagonal matrix and D1⊕F is the direct sum of D1 and F. If we denote Γ=Γ2Γ1 and Π=Π1Π2, we obtain (49)Dn-1=ΠD1-1⊕F-1Γ.Since the last row elements of the matrix Π are 0,1,(Tn-1-Tn)/p2,(Tn-2-Tn-1)/p2,…,(T3-T4)/p2,(T2-T3)/p2, then the elements of last row of Π(D1-1⊕F-1) are given by the following equations: (50)x1′=0,x2′=1p2,x3′=γ3l+T1-Tn+1∑i=2n-2gi1γi+2l,x4′=-γ3∑i=2n-3gi1ρi+3l+∑i=1n-3‍gi1γi+3-T1-Tn+1∑i=1n-3gi1ρi+3∑i=2n-2gi1γi+2l,⋮xk′=-γ3∑i=2n-k+1gi1ρi+k-1l+∑i=1n-k+1‍gi1γi+k-1-T1-Tn+1∑i=1n-k+1gi1ρi+k-1∑i=2n-2gi1γi+2l,k≥4,where (51)γi=Tn-i+2-Tn-i+3p2,i=3,4…,n,ρ3=T1-Tn-Tn-1-p3p2Tn-Tn-1,ρi=Tn-i+1-p3p2Tn-i+3-Tn-i+2i=4,5,…,n.By Lemma 6, if Dn-1=Circ(x1,x2,…,xn), then its last row elements are given by the following equations: (52)x2=-x2′+-1+p3p2x3′-x4′,x3=xn′,x4=xn-1′-xn′,x5=xn-2′-xn-1′-xn′,⋮xk=xn-k+3′-xn-k+4′-xn-k+5′-xn-k+6′5<k≤n,x1=x2′+-1-p3p2x3′-x4′-x5′.Hence, the proof is compeleted.


## 3. Determinant and Inverse of Tibonacci Left Circulant Matrix

In this section, letDn′=LCircT1,T2,…,Tn be a Tibonacci left circulant matrix. By using the obtained conclusions, we give a determinant formula for the matrix Dn′. Afterwards, we discuss the invertibility of the matrix Dn′. The inverse of the matrix Dn′ is also presented. According to Lemma 2 in [15] and Theorems 3, 4, and 7, we can obtain the following theorems.Theorem 8.
LetDn′=LCirc(T1,T2,…,Tn) be a Tribonacci left circulant matrix; then one has (53)det⁡Dn′=-1n-1n-2/2×T1-Tn+∑i=1n-2Δi‍Tn-i+1-Tn-iδ1-∑i=1n-2Δi‍-Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2δ2,where Tn is the nth Tribonacci number.Theorem 9.
LetDn′=LCirc(T1,T2,…,Tn) be a Tribonacci left circulant matrix. If n≠2 and n≠2kπ(arctan(±4ac-b2/-b))-1(k=1,2,…,n-1), then Dn′ is an invertible matrix.Theorem 10.
LetDn′=LCirc(T1,T2,…,Tn) be a Tribonacci left circulant matrix. If Dn is an invertible matrix, then we have (54)Dn′-1=Circx2′+-1-p3p2x3′-x4′-x5′,x3′-x4′-x5′-x6′,…,xn-2′-xn-1′-xn′,xn-1′-xn′,xn′,-x2′+-1+p3p2x3′-x4′,where (55)x1′=0,x2′=1p2,x3′=γ3l+T1-Tn+1∑i=2n-2gi1γi+2l,x4′=-γ3∑i=2n-3gi1ρi+3l+∑i=1n-3‍gi1γi+3-T1-Tn+1∑i=1n-3gi1ρi+3∑i=2n-2gi1γi+2l,⋮xk′=-γ3∑i=2n-k+1gi1ρi+k-1l+∑i=1n-k+1‍gi1γi+k-1-T1-Tn+1∑i=1n-k+1gi1ρi+k-1∑i=2n-2gi1γi+2l,k≥4.


## 4. Determinant and Inverse of Tibonaccig-Circulant Matrix

In this section, letDg,n=g-CircT1,T2,…,Tn be a Tibonacci g-circulant matrix. By using the obtained conclusions, we give a determinant formula for the matrix Dg,n. Afterwards, we discuss the invertibility of the matrix Dg,n. The inverse of the matrix Dg,n is also presented. From Lemmas 3 and 4 in [15] and Theorems 3, 4, and 7, we deduce the following results.Theorem 11.
LetDg,n=g-Circ(T1,T2,…,Tn) be a Tribonacci g-circulant matrix and (n,g)=1; then one has (56)det⁡Dg,n=det⁡Qg·T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-iδ1-∑i=1n-3‍-Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2δ2,where Tn is the nth Tribonacci number.Theorem 12.
LetDg,n=g-Circ(T1,T2,…,Tn) be a Tribonacci g-circulant matrix and (n,g)=1. If n≠2 and n≠2kπ(arctan(±4ac-b2/-b))-1(k=1,2,…,n-1), then Dg,n is an invertible matrix.Theorem 13.
LetDg,n=g-Circ(T1,T2,...,Tn) be a Tribonacci g-circulant matrix and (n,g)=1. If Dn is an invertible matrix, then one has (57)Dg,n-1=Circx2′+-1-p3p2x3′-x4′-x5′,-x2′+-1+p3p2x3′-x4′,xn′,xn-1′-xn′,xn-2′-xn-1′-xn′,…,x3′-x4′-x5′-x6′x2′+-1-p3p2x2′+-1-p3p2x3′-x4′-x5′,QgT,where (58)x1′=0,x2′=1p2,x3′=γ3l+T1-Tn+1∑i=2n-2gi1γi+2l,x4′=-γ3∑i=2n-3gi1ρi+3l+∑i=1n-3‍gi1γi+3-T1-Tn+1∑i=1n-3gi1ρi+3∑i=2n-2gi1γi+2l,⋮xk′=-γ3∑i=2n-k+1gi1ρi+k-1l+∑i=1n-k+1‍gi1γi+k-1-T1-Tn+1∑i=1n-k+1gi1ρi+k-1∑i=2n-2gi1γi+2l,k≥4.


## 5. Conclusion

The related problem of Tribonacci circualnt type matrices is studied in this paper. We not only discuss nonsingularity of Tibonacci circulant type matrices, but also give the explicit determinant and inverse of Tribonacci circulant matrix, Tribonacci left circualant matrix, and Tribonaccig-circualant matrix. Furthermore, according to Theorem 11 in [23] and the result in Theorem 3 in the paper, identities can be easily obtained:(59)1-Tn+1n-c1n+d1n+-TnnL-n-Ln=T1-Tn+∑i=1n-2‍ΔiTn-i+1-Tn-iδ1--Tn+ΔT1-Tn-Tn-1+∑i=1n-3‍Δi+1Tn-i-2δ2,where Tn is the nth Tribonacci number, Ln is the nth generalized Lucas number, and(60)c1=Tn+2-Tn+1+μ12,d1=Tn+2-Tn+1-μ12,μ1=Tn+2-Tn+12-4TnTn+1-1,Δ=-b±b2-4ac2a,δ1=T1-Tn-Tn-1T1-Tn+1n-3+∑i=2n-2‍-11+iT1-Tn+1n-i-2·Tn-i-1det⁡Ai-1,δ2=∑i=1n-2‍-11+iT1-Tn+1n-i-2Tn-i+1-Tn-idet⁡Ai-1,det⁡Ai-1=b+b2-4ac/2i-b-b2-4ac/2ib2-4ac,τ22=4τ1τ3.τ22=4τ1τ3.τ22=4τ1τ3.4ac00b2≠4ac,ib2i-1,b2=4ac.a=T1-Tn+1,b=-Tn-Tn-1,c=-Tn.In addition, we will develop solving the problem in [24–26] by circulant matrices technology.



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*Source: 169726-2015-03-19.xml*