**1. Introduction**

Mersenne numbers are ubiquitous in combinatorics, group theory, chaos, geometry, physics, etc. [1]. They are generated by the following recurrence [2]:

$$M\_{n+1} = 3M\_n - 2M\_{n-1} \qquad \text{where} \quad M\_0 = 0, \ M\_1 = 1, \ n \ge 1; \tag{1}$$

$$M\_{-\left(n+1\right)} = \frac{3}{2}M\_{-n} - \frac{1}{2}M\_{-\left(n-1\right)}\text{ where}\quad M\_0 = 0,\ M\_{-1} = -\frac{1}{2},\ n \ge 1. \tag{2}$$

The Binet formula says that the *n*th Mersenne number *Mn* = 2*n* − 1 [3]. One application we would like to mention is that Nussbaumer [4] applied number theoretical transform closely related to Mersenne number to deal with problems of digital filtering and convolution of discrete signals.

In this paper, we study some basic quantities (determinants and inverses) associated with the periodic tridiagonal Toeplitz matrix with perturbed corners of type 1, which is defined as follows

$$\mathbf{A} = \begin{pmatrix} a\_1 & 2\hbar & 0 & \cdots & 0 & \gamma\_1 \\ 0 & -3\hbar & \ddots & \ddots & & 0 \\ 0 & \hbar & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 2\hbar & 0 \\ & & & & \\ 0 & & \ddots & \ddots & -3\hbar & 2\hbar \\ a\_n & 0 & \cdots & 0 & \hbar & \gamma\_n \end{pmatrix}\_{n \times n},\tag{3}$$

where *α*1, *αn*, *γ*1, *γ<sup>n</sup>*, *h*¯ are complex numbers with *h*¯ = 0. Let ˆ*In* be the *n* × *n* "reverse unit matrix", which has ones along the secondary diagonal and zeros elsewhere. A matrix of the form B := ˆ*In*A<sup>ˆ</sup>*In* is

called a periodic tridiagonal Toeplitz matrix with perturbed corners of type 2, we say that B is induced by A. It is readily seen that A is a periodic tridiagonal Toeplitz matrix with perturbed corners of type 1 if and only if its transpose A*<sup>T</sup>* is a periodic tridiagonal Toeplitz matrix with perturbed corners of type 2.

Tridiagonal matrices appear not only in pure linear algebra, but also in many practical applications, such as, parallel computing [5], computer graphics [6], fluid mechanics [7,8], chemistry [9], and partial differential equations [10–15]. Taking linear hyperbolic equation as an example, some scholars have studied some matrices in discretized partial differential equations. Chan and Jin [16] discussed a linear hyperbolic equation considered by Holmgren and Otto [17] in one-dimensional and two-dimensional cases. Here we restate the linear hyperbolic equation in the two-dimensional case,

$$\frac{\partial u(\mathbf{x}\_{1\prime}, \mathbf{x}\_{2\prime}, t)}{\partial t} + v\_1 \frac{\partial u(\mathbf{x}\_{1\prime}, \mathbf{x}\_{2\prime}, t)}{\partial \mathbf{x}\_1} + v\_2 \frac{\partial u(\mathbf{x}\_{1\prime}, \mathbf{x}\_{2\prime}, t)}{\partial \mathbf{x}\_2} = \mathbf{g}\_{\prime\prime}$$

where 0 < *x*1, *x*2 ≤ 1, *t* > 0, *<sup>u</sup>*(*<sup>x</sup>*1, 0, *t*) = *f*(*<sup>x</sup>*1 − *at*), , *u*(0, *x*1, *t*) = *f*(*<sup>x</sup>*2 − *at*), *<sup>u</sup>*(*<sup>x</sup>*1, *x*2, *t*) = *f*(*<sup>x</sup>*1 + *<sup>x</sup>*2), *g* = (*<sup>v</sup>*1 + *v*2 − *a*)*f* . Here *v*1, *v*2, and *a* are positive constants and *f* is a scalar function with derivative *f* . Denote *s*1, *s*2, *k* as the two spatial steps and time step respectively. For simplicity, assume that *v*1 = *v*2 = *v* and *s*1 = *s*2 = s. The linear hyperbolic equation discretized based on trapezoidal rule in time and center difference in two spaces, respectively. It's coefficient matrix is a tridiagonal matrix with perturbed last row:

$$
\wp = \begin{pmatrix}
\mathbf{2} & \odot & \mathbf{0} & \cdots & \cdots & \cdots & \mathbf{0} \\
\mathbf{0} & \ddots & \ddots & \ddots & \ddots & & & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\
\vdots & & \ddots & \ddots & \ddots & \ddots & \mathbf{0} \\
\vdots & & & \ddots & \ddots & \ddots & \ddots & \mathbf{0} \\
\vdots & & & & \ddots & -\odot & \mathbf{2} & \odot \\
\mathbf{0} & \cdots & \cdots & \cdots & \mathbf{0} & -\mathbf{2}\odot & \mathbf{2} + \mathbf{2}\odot \\
\end{pmatrix}\_{\mathbf{n}\times\mathbf{n}}
$$

,

where = *vk*/s. On the other hand, some parallel computing algorithms are also designed for solving tridiagonal systems on graphics processing unit (GPU), which are parallel cyclic reduction [18] and partition methods [19]. Recently, Yang et al. [20] presented a parallel solving method which mixes direct and iterative methods for block-tridiagonal equations on CPU-GPU heterogeneous computing systems, while Myllykoski et al. [21] proposed a generalized graphics processing unit implementation of partial solution variant of the cyclic reduction (PSCR) method to solve certain types of separable block tridiagonal linear systems. Compared to an equivalent CPU implementation that utilizes a single CPU core, PSCR method indicated up to 24-fold speedups.

On the other hand, many studies have been conducted for tridiagonal matrices or periodic tridiagonal matrices, especially for their determinants and inverses [22–30]. Two decades ago, Wittenburg [31] studied the inverse of tridiagonal toeplitz and periodic matrices and applied them to elastostatics and vibration theory. Recently, El-Mikkawy and Atlan [32] proposed a symbolic algorithm based on the Doolittle LU factorization and Jia et al. put forward some algorithms [33–35] based on block diagonalization technique for *k*-tridiagonal matrix. In 2018, Tim and Emrah [36] used backward continued fractions to derive the LU factorization of periodic tridiagonal matrix and then derived an explicit formula for its inverse. Furthermore, some scholars were attracted by the fact that one could view periodic tridiagonal Toeplitz matrices as a special case of periodic tridiagonal matrices. Shehawey [37] generalized Huang and McColl's [38] work and put forward the inverse formula for periodic tridiagonal Toeplitz matrices.

The rest of the paper is organized as follows: Section 2 describes the detailed derivations of the determinants and inverses of periodic tridiagonal Toeplitz matrices with perturbed corners through matrix transformations, Schur complement and matrix decomposition with the Sherman-Morrison-Woodbury formula [39]. Specifically, the formulas on representation of the determinants and inverses of these typies matrices in the form of products of Mersenne numbers and some initial values. Furthermore, the properties of the periodic tridiagonal Toeplitz matrices with perturbed corners of type 2 can also be obtained. Section 3 presents the numerical results to test the effectiveness of our theoretical results. The final conclusions are given in Section 4.

#### **2. Determinants and Inverses**

In this section, we derive explicit formulas for the determinants and inverses of a periodic tridiagonal Toeplitz matrix with perturbed corners. Main effort is made to work out those for periodic tridiagonal Toeplitz matrix with perturbed corners of type 1, since the results for type 2 matrices would follow immediately.

**Theorem 1.** *Let* A = (*ai*,*j*)*<sup>n</sup> i*,*j*=1 (*n* ≥ 3) *be an n* × *n periodic tridiagonal Toeplitz matrix with perturbed corners of type* 1*. Then*

$$\det \mathbf{A} = (-\hbar)^{n-2} \left\{ \left[ 2M\_{n-2}a\_1 - 4(M\_{n-3} + 1)a\_n \right] \hbar + M\_{n-1}(a\_1\gamma\_n - a\_n\gamma\_1) \right\},\tag{4}$$

*where Mi* (*i* = *n* − 3, *n* − 2, *n* − 1) *is the ith Mersenne number.*

**Proof.** Define the circulant matrix

$$
\epsilon = (\epsilon\_{i,j})\_{i,j=1}^n \tag{5}
$$

where

$$
\varepsilon\_{i,j} = \begin{cases} 1, & i = n, j = 1, \\ 1, & j = i + 1, \\ 0, & \text{otherwise.} \end{cases}
$$

Clearly, is invertible, and

$$\det \mathfrak{e} = (-1)^{n-3}.\tag{6}$$

 (7)

Multiply A by from right and then partition A into four blocks:

$$\mathsf{A}\boldsymbol{\epsilon} = \begin{pmatrix} \gamma\_1 & \alpha\_1 & \vdots & 2\hbar & 0 & \cdots & \cdots & \cdots & 0 \\ \cdot & 0 & -\beta\_1 & -2\hbar & 2\hbar & & & & \\ 0 & 0 & \hbar & -3\hbar & 2\hbar & 0 & & & \\ \vdots & \vdots & \vdots & 0 & \hbar & -3\hbar & 2\hbar & \ddots & \vdots \\ \vdots & \vdots & \vdots & 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \ddots & \ddots & 2\hbar \\ 0 & \vdots & \vdots & \vdots & & \ddots & \ddots & 2\hbar \\ 2\hbar & 0 & \vdots & \vdots & \ddots & \ddots & \ddots & -3\hbar \\ \gamma\_n & \alpha\_n & 0 & 0 & \cdots & \cdots & 0 & \hbar \end{pmatrix}$$
 
$$= \begin{pmatrix} \mathbb{A}\_{11} & \mathbb{A}\_{12} \\ \mathbb{A}\_{21} & \mathbb{A}\_{22} \end{pmatrix}.$$

Since A22 is upper triangular, its determinant is clear which is

$$\det A\_{22} = \hbar^{n-2}.\tag{8}$$

As we assume *h*¯ = 0, so A22 is invertible. It is known (see, e.g., ([29], Lemma 2.5)) that A−<sup>1</sup> 22 = (*a*¨*i*,*j*)*ni*,*j*=<sup>1</sup> where

$$\ddot{a}\_{i,j} = \begin{cases} \frac{M\_{j-i+1}}{\hbar}, & i \le j\_{\prime} \\ 0, & i > j\_{\prime} \end{cases}$$

and *Mi* is the *i*th Mersenne number.

> Next, taking the determinants for both sides of (7) and by (see, e.g., ([40], p. 10)), we ge<sup>t</sup>

$$\det(\mathbb{A}\epsilon) = \det \mathbb{A}\_{22} \det(\mathbb{A}\_{11} - \mathbb{A}\_{12} \mathbb{A}\_{22}^{-1} \mathbb{A}\_{21}).\tag{9}$$

Therefore

$$\det \mathbf{A} = \frac{\det \mathbf{A}\_{22} \det(\mathbf{A}\_{11} - \mathbf{A}\_{12} \mathbf{A}\_{22}^{-1} \mathbf{A}\_{21})}{\det \boldsymbol{\epsilon}}. \tag{10}$$

To find det A, we need to evaluate the determinant of (A11 − <sup>A</sup>12A−<sup>1</sup> 22 A21). From (7) we have

$$\mathbb{A}\_{11} - \mathbb{A}\_{12}\mathbb{A}\_{22}^{-1}\mathbb{A}\_{21} = \begin{pmatrix} \gamma\_1 - 2M\_{n-2}\gamma\_n - 4M\_{n-3}\hbar & \alpha\_1 - 2M\_{n-2}\alpha\_n\\ M\_{n-1}\gamma\_n + 2M\_{n-2}\hbar & M\_{n-1}\alpha\_n \end{pmatrix},$$

and so

$$\det\left(\mathbb{A}\_{11} - \mathbb{A}\_{12}\mathbb{A}\_{22}^{-1}\mathbb{A}\_{21}\right) = \left[4(M\_{n-3} + 1)a\_n - 2M\_{n-2}a\_1\right]\hbar - M\_{n-1}\left(a\_1\gamma\_n - a\_n\gamma\_1\right). \tag{11}$$

Finally, applying (6), (8), and (11) to (10), we ge<sup>t</sup> the determinant of A, which completes the proof.

**Theorem 2.** *Let* A = (*ai*,*j*)*ni*,*j*=<sup>1</sup>(*<sup>n</sup>* ≥ 3) *be a nonsingular periodic tridiagonal Toeplitz matrix with perturbed corners of type* 1*. Then* A−<sup>1</sup> = (*a*˘*i*,*j*)*ni*,*j*=1, *where*

*a* ˘*i*,*j* = ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩2*Mn*−2*h*¯ +*Mn*−1*γ<sup>n</sup> ψ* , *i* = 1, *j* = 1, 4*Mn*−3*h*¯ <sup>−</sup>*γ*1+2*Mn*−2*γ<sup>n</sup> ψ* , *i* = 1, *j* = 2, (*Mn*−2+<sup>1</sup>)*<sup>α</sup><sup>n</sup>* −*ψ* , *i* = 2, *j* = 1, 2*Mn*−3*α*1*h*¯ <sup>+</sup>*Mn*−<sup>2</sup>(*<sup>α</sup>*1*γn*<sup>−</sup>*αnγ*1) −*ψh*¯ , *i* = 2, *j* = 2, <sup>3</sup>(*Mn*−3+<sup>1</sup>)*<sup>α</sup><sup>n</sup>* −*ψ* , *i* = 3, *j* = 1, (*Mn*−3−<sup>1</sup>)*<sup>α</sup>*1*h*¯ +(*Mn*−2+<sup>1</sup>)*<sup>α</sup>nh*¯ <sup>+</sup>*Mn*−<sup>3</sup>(*<sup>α</sup>*1*γn*<sup>−</sup>*αnγ*1) −*ψh*¯ , *i* = 3, *j* = 2, <sup>3</sup>*a*˘*i*,*j*−<sup>1</sup> − <sup>2</sup>*a*˘*i*,*j*−<sup>2</sup> + 1*h*¯ , *i* ∈ {2, <sup>3</sup>}, *j* = *i* + 1, <sup>3</sup>*a*˘*i*,*j*−<sup>1</sup> − <sup>2</sup>*a*˘*i*,*j*−2, *i* ∈ {1, 2, <sup>3</sup>}, *i* + 2 ≤ *j* ≤ *n*; 3 ≤ *j* ≤ *i* ≤ *n*, 32 *<sup>a</sup>*˘*i*−1,*<sup>j</sup>* − 12 *<sup>a</sup>*˘*i*−2,*j*, *j* ∈ {1, <sup>2</sup>}, 4 ≤ *i* ≤ *n*; 4 ≤ *i* < *j* ≤ *n*, (12) *ψ* = 2*Mn*−2*α*1*h*¯− (*Mn*−<sup>1</sup>+ <sup>1</sup>)*<sup>α</sup>nh*¯+ *Mn*−<sup>1</sup>(*<sup>α</sup>*1*γ<sup>n</sup>*− *<sup>α</sup>nγ*1),(13)

*and Mi* (*i* = *n* − 3, *n* − 2, *n* − 1) *is the ith Mersenne number.*

**Proof.** Let A−<sup>1</sup> = (*a*˘*i*,*j*)*ni*,*j*=<sup>1</sup> and the identity matrix *In* = (*ei*,*j*)*ni*,*j*=1, where

$$x\_{i,j} = \begin{cases} 1, \ i = j, \\ 0, \ otherwise. \end{cases} \tag{14}$$

For a nonsingular A,

$$\mathbf{A}^{-1}\mathbf{A} = \mathbf{A}\mathbf{A}^{-1} = I\_{\mathbf{u}}.\tag{15}$$

According to (15), we ge<sup>t</sup>

$$\mathfrak{e}\_{i,j} = 2\mathfrak{h}\_{i,j-1}\hbar - 3\mathfrak{d}\_{i,j}\hbar + \mathfrak{d}\_{i,j+1}\hbar,\tag{16} \le i \le n, \, 2 \le j \le n-1,\tag{16}$$

$$x\_{i,j} = \mathfrak{d}\_{i-1,j}\hbar - 3\mathfrak{d}\_{i,j}\hbar + 2\mathfrak{d}\_{i+1,j}\hbar,\tag{17}$$

Based on (14),we ge<sup>t</sup> from (16) that

$$\mathfrak{A}\_{i,j} = 3\mathfrak{A}\_{i,j-1} - 2\mathfrak{A}\_{i,j-2}, \begin{cases} i \in \{1, 2, 3\}, i+2 \le j \le n; \\ 3 \le j \le i \le n, \end{cases} \tag{18}$$

and *<sup>a</sup>*˘*i*,*i*+<sup>1</sup> = 3*a*˘*i*,*<sup>i</sup>* − <sup>2</sup>*a*˘*i*,*i*−<sup>1</sup> + 1*h*¯ for *i* = 2, 3. Similarly,from (17),wege<sup>t</sup>that

$$\mathfrak{u}\_{i,j} = \frac{3\mathfrak{u}\_{i-1,j}}{2} - \frac{\mathfrak{u}\_{i-2,j}}{2}, \begin{cases} j \in \{1,2\}, 4 \le i \le n; \\ 4 \le i < j \le n. \end{cases} \tag{19}$$

Therefore, based on the above analysis, we need to determine six initial values, that is, *<sup>a</sup>*˘*i*,*<sup>j</sup>* (*i* ∈ {1, 2, <sup>3</sup>}, *j* ∈ {1, <sup>2</sup>}), for the recurrence relations (18) and (19) in order to compute the inverse of A. The rest of the proof is devoted to evaluating these particular entries of A−1.

We decompose A as follows:

$$
\mathbb{A} = \hbar \Delta + F\mathcal{G},
\tag{20}
$$

where  $\Delta = 3T\_{M,n'}^{-1}F = \left(f\_1^T, f\_2^T\right), G = \left(\begin{array}{c} \mathcal{S}1\\\mathcal{S}2 \end{array}\right)$  with 
$$f\_1 = \left(a\_1 + \frac{2M\_n\hbar}{M\_{n+1}}, -\hbar, 0, \dots, 0, a\_n - \frac{2\hbar}{M\_{n+1}}\right)\_{1 \times n'},$$
 
$$f\_2 = \left(\gamma\_1 - \frac{(M\_n + 1)\hbar}{M\_{n+1}}, 0, \dots, 0, \gamma\_n + \frac{2M\_n\hbar}{M\_{n+1}}\right)\_{1 \times n'}$$
 
$$g\_1 = \left(1, 0, \dots, 0\right)\_{1 \times n'}$$
 
$$g\_2 = \left(0, \dots, 0, 1\right)\_{1 \times n'}$$

and *Mi* the *i*th Mersenne number as before.

It could be verified that Δ−<sup>1</sup> = 13 (*tij*)*ni*,*j*=1, where

$$t\_{ij} = \begin{cases} M\_{j-i+1\nu} & 1 \le i \le j \le n\_{\nu} \\ -2M\_{j-i-1\nu} & 1 \le j < i \le n\_{\nu} \end{cases}$$

and *M*−*<sup>m</sup>* is given in (2) for *m* = 1, 2, . . .. Applying the Sherman-Morrison-Woodbury formula (see, e.g., ([39] p. 50)) to (20) gives

$$\mathbf{A}^{-1} = (\hbar \boldsymbol{\Lambda} + \boldsymbol{F} \boldsymbol{G})^{-1} = \frac{1}{\hbar} \boldsymbol{\Lambda}^{-1} - \frac{1}{\hbar^2} \boldsymbol{\Lambda}^{-1} \boldsymbol{F} (\boldsymbol{I}\_n + \frac{1}{\hbar} \boldsymbol{G} \boldsymbol{\Lambda}^{-1} \boldsymbol{F})^{-1} \boldsymbol{G} \boldsymbol{\Lambda}^{-1}.\tag{21}$$

Now we compute each component on the right side of (21). Multiplying respectively Δ−<sup>1</sup> by *G* and *F* from left and right,

$$\mathbf{G}\boldsymbol{\Delta}^{-1} = \frac{1}{3} \begin{pmatrix} \eta\_1 \\ \eta\_2 \\ \vdots \end{pmatrix} \prime \tag{22}$$

$$
\Delta^{-1}F = \frac{1}{3} \left( \begin{array}{cc} \upsharp\_1 & \upsharp\_2 \\ \end{array} \right),
\tag{23}
$$

where *η*1 and *η*2 are row vectors, *ξ*1 and *ξ*2 are column vectors,

$$\begin{aligned} \eta\_1 &= (\mathcal{M}\_{\dot{l}})\_{\dot{j}=1'}^n \\ \eta\_2 &= (-2\mathcal{M}\_{\dot{j}-n-1})\_{\dot{j}=1'}^n \\ \xi\_1^T &= \left(\xi\_{1,1} - 3\hbar \, \zeta\_{2,1} \zeta\_{3,1} \cdot \dots \, \right. \\ \xi\_{i,1} &= \mathcal{M}\_{n-i+1} \alpha\_n - 2\mathcal{M}\_{-i} \alpha\_1 \; \; i = 1, 2, \dots, n, \\ \xi\_2 &= \left(\mathcal{M}\_{n-i+1} \gamma\_n - 2\mathcal{M}\_{-i} \gamma\_1 + 2\mathcal{M}\_{n-i} \hbar \right)\_{i=1}^n. \end{aligned}$$

Then multiplying (23) by *Gh*¯ from the left, further adding *In* and computing the inverse of the matrix

$$\left(I\_{\hbar} + \frac{G}{\hbar} \Delta^{-1} F\right)^{-1} = \frac{3\hbar}{\hbar} \begin{pmatrix} -2M\_{-n}\gamma\_1 + \gamma\_n + 3\hbar & -(\gamma\_1 + M\_n \gamma\_n + 2M\_{n-1}\hbar) \\ 2M\_{-n}a\_1 - a\_n & a\_1 + M\_n a\_n \end{pmatrix},$$

where *h* = *Mn*+<sup>1</sup>)*<sup>M</sup>*1−*<sup>n</sup>*(*<sup>α</sup>*1*γ<sup>n</sup>* − *<sup>α</sup>nγ*1) + *M*2−*nα*1*h*¯ − *<sup>α</sup>nh*¯ \*. Multiplying the pervious formula -*In* + *Gh*¯ <sup>Δ</sup>−1*F*−<sup>1</sup> by Δ−1*F* from the left and by *G*Δ−<sup>1</sup> from the right, respectively, yields

$$
\Delta^{-1} F \left( I\_n + \frac{1}{\hbar} G \Delta^{-1} F \right)^{-1} G \Delta^{-1} = (k\_{\vec{l}\vec{j}})^n\_{\vec{l}, \vec{j} = 1 \prime} \tag{24}
$$

where

$$\begin{aligned} k\_{1j} &= \frac{\theta\_j' \hbar^3 + (\theta\_j'' \gamma\_1 + \theta\_j''' \gamma\_n) \hbar^2}{M\_{n+1} \psi} + \frac{M\_j \hbar}{3}, & 1 \le j \le n\_r, \\\ k\_{1r} &= (a\_1 \eta\_{ij}' + a\_n \eta\_{ij}'') \hbar^2 + (a\_1 \gamma\_n - a\_n \gamma\_1) \eta\_{ij}''' \hbar \\\ &\quad \gamma\_{r-1, r-n\_r+1, \dots, n\_r} \end{aligned}$$

$$\begin{aligned} k\_{ij} &= \frac{\omega\_{n-1,j} + \omega\_{n-1,j} + \dots + \omega\_{n,j}}{3M\_{n-1}\psi}, & 2 \le i \le n, 1 \le j \le n, \\\ \psi &= 2M\_{n-2}n\_1\hbar - (M\_{n-1} + 1)a\_n\hbar + M\_{n-1}(a\_1\gamma\_n - a\_n\gamma\_1), \\\ \psi' &= 2M\_{n-2}(M\_{n-1} + 1) - M\_{n-1}\gamma\_{n-1,j} \end{aligned}$$

$$\theta\_j' = 3M\_{\hat{j}}(M\_{n-1} + 1) - M\_{n-1}M\_{n-j+1}(M\_{\hat{j}} + 1), \tag{1}$$

$$\theta\_j^{\prime\prime} = M\_\text{n} M\_\text{j} - M\_{n-j+1} (M\_{j-1} + 1), \tag{1 \le j \le n\_\text{th}}$$

$$\begin{aligned} \theta\_j^{\prime\prime} &= M\_j(M\_{n-1}+1) - M\_n M\_{n-j+1}(M\_{j-1}+1), & 1 \le j \le n, \\\eta\_{ij}^{\prime} &= 2M\_n M\_{n-i} M\_j - 3M\_i M\_j (M\_{n-i}+1) + M\_n M\_{i-1} M\_{n-j+1} (M\_{j-i+1}+1), & 2 \le i \le n, 1 \le j \le n, \\\eta\_{ij}^{\prime\prime} &= M\_{i-1} M\_{n+j-1} (M\_{n+j-1}+1) - M\_{n-i+2} M\_j (M\_{n-1}+1), & 2 \le i \le n, 1 \le j \le n, \\\eta\_{ij}^{\prime\prime} &= M\_{n+1} [M\_{n-i} M\_j + M\_{i-1} M\_{n-j+1} (M\_{j-i}+1)], & 2 \le i \le n, 1 \le j \le n. \end{aligned}$$

> From (21) and (24), we have

$$(\mathfrak{d}\_{i,j})\_{i,j=1}^{n} = \frac{1}{\hbar} \Delta^{-1} - \frac{1}{\hbar^2} (k\_{i\bar{j}})\_{i,j=1}^{n} \tag{25}$$

where

$$\mathbb{1}\_{i,j} = \frac{M\_{j-i+1}}{3\hbar} - \frac{k\_{i,j}}{\hbar^2}, \tag{26}$$

$$\mathfrak{A}\_{i,j} = -\frac{2M\_{j-i-1}}{3\hbar} - \frac{k\_{i,j}}{\hbar^2}, \tag{27}$$

By (26) we compute,

$$\begin{aligned} \mathfrak{A}\_{1,1} &= \frac{2M\_{\mathfrak{n}-2}\hbar + M\_{\mathfrak{n}-1}\gamma\_{\mathfrak{n}}}{\psi}, \\ \mathfrak{A}\_{1,2} &= \frac{4M\_{\mathfrak{n}-3}\hbar - \gamma\_{1} + 2M\_{\mathfrak{n}-2}\gamma\_{\mathfrak{n}}}{\psi}, \\ \mathfrak{A}\_{2,2} &= \frac{2M\_{\mathfrak{n}-3}\mathfrak{a}\_{1}\hbar + M\_{\mathfrak{n}-2}(\mathfrak{a}\_{1}\gamma\_{\mathfrak{n}} - \mathfrak{a}\_{\mathfrak{n}}\gamma\_{1})}{-\psi\hbar}. \end{aligned}$$

By (27) we compute,

$$\begin{aligned} d\_{2,1} &= \frac{(M\_{n-2} + 1)a\_n}{-\psi}, \\ d\_{3,1} &= \frac{3(M\_{n-3} + 1)a\_n}{-\psi}, \\ d\_{3,2} &= \frac{(M\_{n-3} - 1)a\_1\hbar + (M\_{n-2} + 1)a\_n\hbar + M\_{n-3}(a\_1\gamma\_{\mathcal{U}} - a\_n\gamma\_1)}{-\psi\hbar}. \end{aligned}$$

This completes the proof.

**Remark 1.** *Formulas* (26) *and* (27) *would give an analytic formula for* A−1*. However, there is a big advantage of* (12) *from computational consideration as we shall see from Section 3.*

The next two theorems are parallel results for type 1 matrices.

**Theorem 3.** *Let* A *be a periodic tridiagonal Toeplitz matrix with perturbed corners of type* 1 *and* B *be a periodic tridiagonal Toeplitz matrix with perturbed corners of type* 2*, which is induced by* A*. Then*

$$\det \mathbb{B} = (-\hbar)^{n-2} \left\{ \left[ 2M\_{n-2}\alpha\_1 - 4(M\_{n-3} + 1)\alpha\_n \right] \hbar + M\_{n-1}(\alpha\_1 \gamma\_n - \alpha\_n \gamma\_1) \right\}.$$

**Proof.** Since detB = det ˆ *In* det A det ˆ*In*, we obtain this conclusion by using Theorem 1 and det ˆ*In* = (−<sup>1</sup>) *<sup>n</sup>*(*<sup>n</sup>*−<sup>1</sup>) 2 .

**Theorem 4.** *Let* A *be a periodic tridiagonal Toeplitz matrix with perturbed corners of type* 1 *and* B *be a periodic tridiagonal Toeplitz matrix with perturbed corners of type* 2*, which is induced by* A*. Then*

$$\mathbb{B}^{-1} = (\mathfrak{A}\_{n+1-i\mu+1-j})\_{i,j=1}^n$$

*where <sup>a</sup>*˘*i*,*<sup>j</sup> is the same as* (12)*.*

**Proof.** It follows immediately from B−<sup>1</sup> = ˆ*I*−<sup>1</sup> *n* A−<sup>1</sup> ˆ*I*−<sup>1</sup> *n* = ˆ*In*A−<sup>1</sup> ˆ*In* and Theorem 2.
