Square Integer Matrix with a Single Non-Integer Entry in Its Inverse

Abstract

Matrix inversion is one of the most significant operations on a matrix. For any non-singular matrix $A\in {\mathbb{Z}}^{n\times n}$, the inverse of this matrix may contain countless numbers of non-integer entries. These entries could be endless floating-point numbers. Storing, transmitting, or operating such an inverse could be cumbersome, especially when the size n is large. The only square integer matrix that is guaranteed to have an integer matrix as its inverse is a unimodular matrix $U\in {\mathbb{Z}}^{n\times n}$. With the property that $det(U)=\pm 1$, then $U^{-1}\in {\mathbb{Z}}^{n\times n}$ is guaranteed such that $U{U}^{-1}=I$, where $I\in {\mathbb{Z}}^{n\times n}$ is an identity matrix. In this paper, we propose a new integer matrix $\tilde{G}\in {\mathbb{Z}}^{n\times n}$, which is referred to as an almost-unimodular matrix. With $det(\tilde{G})\ne \pm 1$, the inverse of this matrix, ${\tilde{G}}^{-1}\in {\mathbb{R}}^{n\times n}$, is proven to consist of only a single non-integer entry. The almost-unimodular matrix could be useful in various areas, such as lattice-based cryptography, computer graphics, lattice-based computational problems, or any area where the inversion of a large integer matrix is necessary, especially when the determinant of the matrix is required not to equal $\pm 1$. Therefore, the almost-unimodular matrix could be an alternative to the unimodular matrix.

1. Introduction

In some methods, the inversion of an integer matrix is an avoidable operation. For example, Babai’s Round-off Method (RoM) is an approximation method that was proposed by Babai [1]. This method helps approximate the Closest-Vector Problem (CVP) [2], which is one of the most established lattice-based computational problems. For instance, Babai’s RoM is used in the decryption algorithm of the lattice-based Goldreich–Goldwasser–Halevi encryption scheme (GGH crypto-system) [3] and its upgraded version [4]. The inversion of a large integer matrix in the decryption algorithms of these schemes becomes a significant issue that might trigger decryption errors.

For $n\in \mathbb{N}$, consider $\mathcal{L}\subset {\mathbb{R}}^n$ as a full-rank lattice. Suppose that $B=\{{\overrightarrow{b}}_1,{\overrightarrow{b}}_2,\cdots ,{\overrightarrow{b}}_n\}$, where ${\overrightarrow{b}}_1,{\overrightarrow{b}}_2,\cdots ,{\overrightarrow{b}}_n\in {\mathbb{R}}^n$ is a basis for the lattice $L(B)=\mathcal{L}\subset {\mathbb{R}}^n$. To approximate the CVP in the lattice $L(B)$, the input of Babai’s RoM is the lattice basis B. For that purpose, this basis is represented in matrix form as $B\in {\mathbb{R}}^{n\times n}$, where ${\overrightarrow{b}}_1,{\overrightarrow{b}}_2,\cdots ,{\overrightarrow{b}}_n$ become the columns of the matrix B. One of the crucial steps in this method involves the inversion of the lattice basis B. For the purpose of computational efficiency, the basis B is generated as an integer matrix $B\in {\mathbb{Z}}^{n\times n}$ [3].

Nevertheless, the inverse $B^{-1}\in {\mathbb{R}}^n$ is not necessarily an integer matrix. The non-integer entries of the matrix $B^{-1}$ could be long floating-point numbers. For small dimensions, n, the inversion of the basis B is efficiently performed, and storing all entries, including the non-integer entries in the matrix $B^{-1}$, requires a low storage capacity. However, these tasks could become cumbersome once the dimension n becomes larger, such as $n\ge 500$. The matrix $B^{-1}$ may consist of countless numbers of non-integer entries in the form of long floating-point numbers. Storing, transmitting, or operating such a matrix would require expensive computational costs [5]. In other methods where the inversion of a large integer matrix is necessary, the same issues could also arise, and this drawback may limit the true potential of the method, either in terms of its performance or precision. The only square integer matrix that is guaranteed to have an integer inverse is a unimodular matrix $U\in {\mathbb{Z}}^{n\times n}$, where $det(U)=\pm 1$ [6]. With this property, it can be guaranteed that $U^{-1}\in {\mathbb{Z}}^{n\times n}$ such that $U{U}^{-1}=I$, where $I\in {\mathbb{Z}}^{n\times n}$ is an identity matrix.

Hanson introduced a new class of unimodular matrices—dubbed nice matrices—in 1982. These matrices are also invertible, with an inverse comprising entirely integer elements [7]. The nice matrices are triangular in shape, with their diagonal elements having a product equal to $\pm 1$. As a result, they have a determinant equal to $\pm 1$. Nice matrices are also considered unimodular matrices due to their features. Hanson later presented another class of unimodular matrices in 1985, dubbed very nice matrices. Assume that the matrix $A\in {\mathbb{Z}}^{n\times n}$ is a nice matrix. If A’s inverse equals itself, it is classified as a very nice matrix [8]. Nonetheless, the very nice matrix is likewise a unimodular matrix with determinant $\pm 1$. The primary difference between Hansen’s classes and the set of all unimodular matrices is that Hansen’s classes need a square and triangular shape, but the set of all unimodular matrices does not. Although inversion of an integer matrix is critical, there are just a few excellent publications in the literature that are not Hanson’s. Most of the discussion on the inversion of an integer matrix directly relates to unimodular matrices [9,10,11,12].

This paper proposes a new integer matrix that is referred to as an almost-unimodular matrix and is denoted as $\tilde{G}\in {\mathbb{Z}}^{n\times n}$. Compared to the unimodular matrix, the determinant of the almost-unimodular matrix is not equal to $\pm 1$ (i.e., $det(\tilde{G})\ne \pm 1$). The inverse of the almost-unimodular matrix is proven to consist of only a single non-integer entry, regardless of how large the dimension n is. With this property, the almost-unimodular matrix $\tilde{G}$ might be useful in any method or technique where the inversion operation of a large integer matrix is necessary, especially when the determinant of the matrix is required not to equal $\pm 1$.

2. Mathematical Foundation

This section provides some mathematical foundations related to the inversion of the integer matrix. In further discussions, let $n,i,j,k,l\in \mathbb{N}$. Now, consider the following definitions regarding the matrix inversion operation.

Definition 1.

[13] For $n\ge 2$, let $G\in {\mathbb{Z}}^{n\times n}$ with entries $g_{i,j}\in \mathbb{Z}$ and $G_{i,j}\in {\mathbb{Z}}^{(n-1)\times (n-1)}$ denoting the sub-matrix of G that is obtained by deleting the i-th row and j-th column of G. The determinant of G, denoted as $det(G)\in \mathbb{Z}$, is $det(G)={\sum }_{j=1}^n{(-1)}^{i+j}{g}_{i,j}det{G}_{i,j}$.

Using the determinant of the sub-matrices $G_{i,j}$, the $(i,j)$-minor of G, denoted as $m_{i,j}(G)\in \mathbb{Z}$, can be obtained. Then, the $(i,j)$-cofactor of G, denoted as $c_{i,j}(G)\in \mathbb{Z}$, and the $(i,j)$-adjoint of G, denoted as $a_{i,j}(G)\in \mathbb{Z}$, can be determined for all $i,j=1,\dots ,n$.

Definition 2.

[13] Let $G\in {\mathbb{Z}}^{n\times n}$; then, $M(G)$ denotes the minor matrix of G with entries $m_{i,j}(G)\in \mathbb{Z}$, $C(G)$ denotes the cofactor matrix of G with entries $c_{i,j}(G)\in \mathbb{Z}$, and $Adj(G)$ denotes the adjoint matrix of G with entries $a_{i,j}(G)\in \mathbb{Z}$, where $c_{i,j}(G)={(-1)}^{i+j}{m}_{i,j}(G)$. Since $Adj(G)=C{(G)}^T$, then $a_{i,j}(G)=c_{j,i}(G)$ for all $i,j=1,\dots ,n$.

Consider the following definition regarding a diagonal matrix.

Definition 3.

Let $\alpha ,\beta ,\gamma \in \mathbb{Z}$ and $n\ge 3$. $D_n\in {\mathbb{Z}}^{n\times n}$ with $D_n=diag(\alpha ,\beta ,\cdots ,\beta )$ is a diagonal matrix with entries $d_{1,1}=\alpha$ and $d_{i,i}=\beta$ for all $i=2,\dots ,n$ and $D_n^{*}\in {\mathbb{Z}}^{n\times n}$ with $D_n^{*}=diag(\alpha ,\cdots ,\alpha ,\beta )$ is a diagonal matrix with entry $d_{i,i}=\alpha$ for all $i=1,\dots ,n-1$ and $d_{n,n}=\beta$. Moreover, $D_{\alpha ,\beta }\in {\mathbb{Z}}^{n\times n}$ is a diagonal matrix with entry $d_{i,j}\in \{\alpha ,\beta \}$ in a random position and $D_{\alpha ,\beta ,\gamma }\in {\mathbb{Z}}^{n\times n}$ is a diagonal matrix with entry $d_{i,j}\in \{\alpha ,\beta ,\gamma \}$ in a random position for all $i,j=1,\dots ,n$. For the matrix $D_{\alpha ,\beta }$, at least one of the entries (α or β) appears in its diagonal. In contrast, for the matrix $D_{\alpha ,\beta ,\gamma }$, at least one of the entries ($\alpha ,\beta$, or γ) appears in its diagonal.

For any matrix $G\in {\mathbb{Z}}^{n\times n}$, the inverse of this matrix, $G^{-1}\in {\mathbb{R}}^{n\times n}$, is not necessarily an integer matrix. It may contain various non-integer entries. However, there are integer matrices with the property that the inverses of these matrices are guaranteed integer matrices. Such matrices are referred to as unimodular matrices, $U\in {\mathbb{Z}}^{n\times n}$, where $det(U)=\pm 1$. According to Uhlmann and Wang [14], it is precious in the case of $n\times n$ matrices to be able to demonstrate that the result of a given matrix function or transformation has a particular property, such as being unitary or unimodular. These special types of matrices can be used for analysis or to obtain solutions more efficiently compared to general matrices. The importance of unimodular matrices is also discussed in [15,16], where they constitute vital components of lattices in lattice-based cryptography. This is due to their practicality in computations; thus, they have given rise to many studies related to unimodular matrices. The following section explains a method for constructing integer matrices that have some of these desirable properties.

3. Construction of the Almost-Unimodular Matrix

This section discusses the construction of the proposed matrix $\tilde{G}\in {\mathbb{Z}}^{n\times n}$, which we call an almost-unimodular matrix $\tilde{G}$. To ensure that the almost-unimodular matrix is a non-singular and non-unimodular matrix, it is designed in such a way that $det(\tilde{G})\ne 0,\pm 1$. The almost-unimodular matrix $\tilde{G}$ is formed through a multiplication of two triangular matrices $G_1,G_2\in {\mathbb{Z}}^{n\times n}$. In the following discussion, let $I_n$ denote an $n\times n$-identity matrix, $H_U\in {\mathbb{Z}}^{n\times n}$ denote a strictly upper-triangular matrix with upper-triangular entries $h_{i,j}^U\in \mathbb{Z}$, $H_L\in {\mathbb{Z}}^{n\times n}$ denote a strictly lower-triangular matrix with lower-triangular entries $h_{i,j}^L\in \mathbb{Z}$, and $H_{LU}=H_L+H_U$ for all $i\ne j$ and $i,j=1,\cdots ,n$. Before we present the construction of the matrix $\tilde{G}$, consider the following theorem.

Theorem 1.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $T_U,T_L\in {\mathbb{Z}}^{n\times n}$, where $T_U=I_n+\delta S_U$, $T_L=I_n+\delta S_L$, and $S_U,S_L\in {\mathbb{Z}}^{n\times n}$. Then, the minor matrix of $T_U$ is $M(T_U)=I_n+\delta V_L$ and the minor matrix of $T_L$ is $M(T_L)=I_n+\delta V_U$, where $V_U,V_L\in {\mathbb{Z}}^{n\times n}$.

Proof. 

Consider the matrix $T_U$. For all $i=j=1,\cdots ,n$, we have

$$m_{i,j}(T_U)=\left|I_{n-1}+\delta S_U^{\prime }\right|$$

where $S_U^{\prime }\in {\mathbb{Z}}^{(n-1)\times (n-1)}$. Since $I_{n-1}+\delta S_U^{\prime }$ is an upper-triangular matrix with 1s as diagonal entries, then $\left|I_{n-1}+\delta S_U^{\prime }\right|=1$. Hence,

$$m_{i,j}(T_U)=1.$$

Then, for all $i<j$ where $i,j=1,\cdots ,n$, we have

$$m_{i,j}(T_U)=\left|D_{0,1}+\delta S_U^{\prime }\right|$$

where $D_{0,1}\in {\mathbb{Z}}^{(n-1)\times (n-1)}$, with at least one diagonal entry that is 0. Since $D_{0,1}+\delta S_U^{\prime }$ is an upper-triangular matrix with a diagonal entry containing 0, then $\left|D_{0,1}+\delta S_U^{\prime }\right|=0$. Hence,

$$m_{i,j}(T_U)=0.$$

Finally, for all $i>j$ where $i,j=1,\dots ,n$, we have

$$m_{i,j}(T_U)=\sum _{k=1}^{i-j}{\delta }^k{s}_k$$

where ${\delta }^k,s_k\in \mathbb{Z}$ for all $k=1,\dots ,i-j$. This implies that $m_{i,j}(T_U)=\delta v_{i,j}^L$, where $v_{i,j}^L\in \mathbb{Z}$ for all $i>j$ and $i,j=1,\dots ,.n$. Therefore,

$$M(T_U)=I_n+\delta V_L$$

where $V_L\in {\mathbb{Z}}^{n\times n}$ contains $v_{i,j}^L$ as lower-triangular entries for all $i>j$ where $i,j=1,\dots ,n$. For the minor matrix of $T_L$, the proof can be carried out by using a similar approach. □

For the construction of the upper-triangular matrix $G_1$, consider the following.

Definition 4.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $D_n=diag\left(\delta ,1,\dots ,1\right)$. Then, $G_1\in {\mathbb{Z}}^{n\times n}$ is defined as $G_1=D_n+\delta P_U$, where $P_U\in {\mathbb{Z}}^{n\times n}$.

Lemma 1.

Let $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$. The minor matrix of $G_1$ is $M(G_1)=D_n^{\prime }+\delta Q_L$, where $D_n^{\prime }=diag\left(1,\delta ,\dots ,\delta \right)$ and $Q_L\in {\mathbb{Z}}^{n\times n}$.

Proof. 

For $i=j=1$, we have

$$m_{1,1}(G_1)=\left|I_{n-1}+\delta P_U^{\prime }\right|$$

where $P_U^{\prime }\in {\mathbb{Z}}^{(n-1)\times (n-1)}$. Since $I_{n-1}+\delta P_U^{\prime }$ is an upper-triangular matrix with diagonal entries 1, then $\left|I_{n-1}+\delta P_U^{\prime }\right|=1$. Hence,

$$m_{1,1}(G_1)=1.$$

Then, for all $i=j$ and $2\le i,j\le n$, we have

$$m_{i,j}(G_1)=\left|D_{n-1}+\delta P_U^{\prime }\right|$$

where $D_{n-1}=diag(\delta ,1,\dots ,1)$. Since $D_{n-1}+\delta P_U^{\prime }$ is an upper-triangular matrix with diagonal entries containing 1s and a $\delta$, then $\left|D_{n-1}+\delta P_U^{\prime }\right|=\delta$. Hence,

$$m_{i,j}(G_1)=\delta .$$

Next, for all $i<j$ where $i,j=1,\dots ,n$, we have

$$m_{i,j}(G_1)=\left|D_{\delta ,0,1}+P_U^{\prime }\right|$$

where $D_{\delta ,0,1}\in {\mathbb{Z}}^{(n-1)\times (n-1)}$ with at least one diagonal entry of 0. Since $D_{\delta ,0,1}+P_U^{\prime }$ is an upper-triangular matrix with diagonal entries containing 0, then

$$m_{i,j}(G_1)=0.$$

Finally, for all $i>j$ where $i,j=1,\dots ,n$, we have

$$m_{i,j}(G_1)=\sum _{k=1}^{i-j}{\delta }^k{p}_k$$

where ${\delta }^k,p_k\in \mathbb{Z}$ for all $k=1,\dots ,i-j$. This implies that

$$m_{i,j}(G_1)=\delta q_{i,j}^L.$$

Therefore,

$$M(G_1)=D_n^{\prime }+\delta Q_L$$

where $Q_L\in {\mathbb{Z}}^{n\times n}$ contains lower-triangular entries $\delta q_{i,j}^L$ for all $i>j$ where $i,j=1,\dots ,n$. □

Remark 1.

Since the matrix $G_1$ is an upper-triangular matrix with a δ and 1s as its diagonal entries, then $det(G_1)=\delta$. By setting $\delta \ne 0,\pm 1$, then $det(G_1)\ne 0,\pm 1$. This implies that the matrix $G_1$ is a non-singular and non-unimodular matrix.

Consider the following example:

Example 1.

For $n=4$, let $G_1\in {\mathbb{Z}}^{4\times 4}$, where $G_1=D_4+\delta P_U$, $D_4=diag(\delta ,1,\dots 1)$, and $P_U\in {\mathbb{Z}}^{4\times 4}$. Thus,

$$G_1=\left[\begin{array}{cccc}\delta & 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]+\delta \left[\begin{array}{cccc}0& p_{1,2}^U& p_{1,3}^U& p_{1,4}^U\\ 0& 0& p_{2,3}^U& p_{2,4}^U\\ 0& 0& 0& p_{3,4}^U\\ 0& 0& 0& 0\end{array}\right]=\left[\begin{array}{cccc}\delta & \delta p_{1,2}^U& \delta p_{1,3}^U& \delta p_{1,4}^U\\ 0& 1& \delta p_{2,3}^U& \delta p_{2,4}^U\\ 0& 0& 1& \delta p_{3,4}^U\\ 0& 0& 0& 1\end{array}\right].$$

Note that

$$M\left(G_1\right)=\left[\begin{array}{cccc}1& 0& 0& 0\\ \delta q_{2,1}^L& \delta & 0& 0\\ \delta q_{3,1}^L& \delta q_{3,2}^L& \delta & 0\\ \delta q_{4,1}^L& \delta q_{4,2}^L& \delta q_{4,3}^L& \delta \end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& \delta & 0& 0\\ 0& 0& \delta & 0\\ 0& 0& 0& \delta \end{array}\right]+\delta \left[\begin{array}{cccc}0& 0& 0& 0\\ q_{2,1}^L& 0& 0& 0\\ q_{3,1}^L& q_{3,2}^L& 0& 0\\ q_{4,1}^L& q_{4,2}^L& q_{4,3}^L& 0\end{array}\right].$$

Therefore, $M(G_1)=D_n^{\prime }+\delta Q_L$.

For the construction of the lower-triangular matrix $G_2$, consider the following.

Definition 5.

Let $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$. Then, $G_2\in {\mathbb{Z}}^{n\times n}$ is defined as a unimodular matrix $G_2=I_n+\delta P_L$, where $P_L\in {\mathbb{Z}}^{n\times n}$.

Lemma 2.

Let $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$. The minor matrix of $G_2$ is $M(G_2)=I_n+\delta Q_U$, where $Q_U\in {\mathbb{Z}}^{n\times n}$.

Proof. 

Note that $G_2=I_n+\delta P_L$, where $P_L\in {\mathbb{Z}}^{n\times n}$. Based on Theorem 1, we have $M(G_2)=I_n+\delta Q_U$. □

Consider the following example.

Example 2.

For $n=4$, let $G_2\in {\mathbb{Z}}^{4\times 4}$, where $G_2=I_4+\delta P_L$ and $P_L\in {\mathbb{Z}}^{4\times 4}$. Thus,

$$G_2=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]+\delta \left[\begin{array}{cccc}0& 0& 0& 0\\ p_{2,1}^L& 0& 0& 0\\ p_{3,1}^L& p_{3,2}^L& 0& 0\\ p_{4,1}^L& p_{4,2}^L& p_{4,3}^L& 0\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ \delta p_{2,1}^L& 1& 0& 0\\ \delta p_{3,1}^L& \delta p_{3,2}^L& 1& 0\\ \delta p_{4,1}^L& \delta p_{4,2}^L& \delta p_{4,3}^L& 1\end{array}\right].$$

Note that

$$M\left(G_2\right)=\left[\begin{array}{cccc}1& \delta q_{1,2}^U& \delta q_{1,3}^U& \delta q_{1,4}^U\\ 0& 1& \delta q_{2,3}^U& \delta q_{2,4}^U\\ 0& 0& 1& \delta q_{3,4}^U\\ 0& 0& 0& 1\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]+\delta \left[\begin{array}{cccc}0& q_{1,2}^U& q_{1,3}^U& q_{1,4}^U\\ 0& 0& q_{2,3}^U& q_{2,4}^U\\ 0& 0& 0& q_{3,4}^U\\ 0& 0& 0& 0\end{array}\right].$$

Therefore, $M(G_2)=I_4+\delta Q_U$.

The almost-unimodular matrix $\tilde{G}$ is constructed as follows.

Definition 6.

Let $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$. Then, $\tilde{G}\in {\mathbb{Z}}^{n\times n}$ is defined as an almost-unimodular matrix $\tilde{G}=G_1{G}_2$.

Lemma 3.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $\tilde{G}\in {\mathbb{Z}}^{n\times n}$ be an almost-unimodular matrix. Then, $det(\tilde{G})=\delta$.

Proof. 

Note that both $G_1,G_2\in {\mathbb{Z}}^{n\times n}$ are triangular matrices where $det\left(G_1\right)=\delta$ and $det\left(G_2\right)=1$. Therefore,

$$det(\tilde{G})=det(G_1{G}_2)=det(G_1)det(G_2)=\delta (1)=\delta .$$

□

Since $det\left(G_1\right)=\delta$, where $\delta \ne 0,\pm 1$, then the almost-unimodular matrix $\tilde{G}$ is a non-singular and non-unimodular matrix.

Lemma 4.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $\tilde{G}\in {\mathbb{Z}}^{n\times n}$ be an almost-unimodular matrix. Then, $\tilde{G}=D_n+D_n^{*}+\delta P_{LU}$, where $D_n=diag\left(\delta ,1,\cdots ,1\right)$ and $D_n^{*}=diag\left(\delta r_{1,1}^{\prime },\cdots ,\delta r_{n-1,n-1}^{\prime },0\right)$.

Proof. 

Note that, $G_1=D_n+\delta P_U$ and $G_2=I_n+\delta P_L$. Thus,

$$\tilde{G}=G_1{G}_2=D_n+\delta P_U+D_n\delta P_L+{\delta }^2{P}_U{P}_L.$$

Since $D_n\delta P_L=\delta P_L$, then

$$\tilde{G}=D_n+{\delta }^2{P}_U{P}_L+\delta P_U+\delta P_L.$$

Note that

$$\begin{array}{cc}\hfill {\delta }^2{P}_U{P}_L& =\left[\begin{array}{ccccc}{\delta }^2{r}_{1,1}& {\delta }^2{r}_{1,2}& \cdots & {\delta }^2{r}_{1,n-1}& 0\\ {\delta }^2{r}_{2,1}& {\delta }^2{r}_{2,2}& \cdots & {\delta }^2{r}_{2,n-1}& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ {\delta }^2{r}_{n-1,1}& {\delta }^2{r}_{n-1,2}& \cdots & {\delta }^2{r}_{n-1,n-1}& 0\\ 0& 0& \cdots & 0& 1\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ccccc}\delta r_{1,1}^{\prime }& \delta r_{1,2}^{\prime }& \cdots & \delta r_{1,n-1}^{\prime }& 0\\ \delta r_{2,1}^{\prime }& \delta r_{2,2}^{\prime }& \cdots & \delta r_{2,n-1}^{\prime }& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \delta r_{n-1,1}^{\prime }& \delta r_{n-1,2}^{\prime }& \cdots & \delta r_{n-1,n-1}^{\prime }& 0\\ 0& 0& \cdots & 0& 1\end{array}\right]\hfill \end{array}$$

where $r_{i,j}^{\prime }=\delta \left(r_{i,j}\right)\in \mathbb{Z}$ for all $i,j=1,\cdots ,n$. Thus,

$$D_n+{\delta }^2{P}_U{P}_L=\left[\begin{array}{ccccc}\delta +\delta r_{1,1}^{\prime }& \delta r_{1,2}^{\prime }& \cdots & \delta r_{1,n-1}^{\prime }& 0\\ \delta r_{2,1}^{\prime }& 1+\delta r_{2,2}^{\prime }& \cdots & \delta r_{2,n-1}^{\prime }& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \delta r_{n-1,1}^{\prime }& \delta r_{n-1,2}^{\prime }& \cdots & 1+\delta r_{n-1,n-1}^{\prime }& 0\\ 0& 0& \cdots & 0& 1\end{array}\right].$$

Hence, the matrix $\tilde{G}$ has the following entries:

$$\begin{array}{cc}\hfill \tilde{G}& =(D_n+{\delta }^2{P}_U{P}_L)+\delta (P_U+P_L)\hfill \\ \hfill & =\left[\begin{array}{ccccc}\delta +\delta r_{1,1}^{\prime }& \delta (r_{1,2}+p_{1,2})& \cdots & \delta (r_{1,n-1}+p_{1,n-1})& \delta p_{1,n}\\ \delta (r_{2,1}+p_{2,1})& 1+\delta r_{2,2}^{\prime }& \cdots & \delta r_{2,n-1}& \delta p_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \delta (r_{n-1,1}+p_{n-1,1})& \delta r_{n-1,2}& \cdots & 1+\delta r_{n-1,n-1}^{\prime }& \delta p_{n-1,n}\\ \delta p_{n,1}& \delta p_{n,2}& \cdots & \delta p_{n,n-1}& 1\end{array}\right].\hfill \end{array}$$

Since $r_{i,j},p_{i,j}\in \mathbb{Z}$, then $r_{i,j}+p_{i,j}=x_{i,j}\in \mathbb{Z}$ for all $i,j=1,\cdots ,n$. Thus,

$$\begin{array}{cc}\hfill \tilde{G}& =\left[\begin{array}{ccccc}\delta & 0& \cdots & 0& 0\\ 0& 1& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 1& 0\\ 0& 0& \cdots & 0& 1\end{array}\right]+\left[\begin{array}{ccccc}\delta r_{1,1}^{\prime }& 0& \cdots & 0& 0\\ 0& \delta r_{2,2}^{\prime }& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & \delta r_{n-1,n-1}^{\prime }& 0\\ 0& 0& \cdots & 0& 0\end{array}\right]\hfill \\ \hfill & +\delta \left[\begin{array}{ccccc}0& x_{1,2}& \cdots & x_{1,n-1}& x_{1,n}\\ x_{2,1}& 0& \cdots & x_{2,n-1}& x_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ x_{n-1,1}& x_{n-1,2}& \cdots & 0& x_{n-1,n}\\ x_{n,1}& x_{n,2}& \cdots & x_{n,n-1}& 0\end{array}\right]\hfill \\ \hfill & =D_n+D_n^{*}+\delta P_{LU}.\hfill \end{array}$$

□

Theorem 2.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $\tilde{G}\in {\mathbb{Z}}^{n\times n}$ be an almost-unimodular matrix. Then, the cofactor matrix of $\tilde{G}$ is $C(\tilde{G})={(-1)}^{i+j}M(\tilde{G})$, where $M(\tilde{G})=D_n+\delta Y_{LU}$ with $D_n=diag(1,\delta y_{2,2},\cdots ,\delta y_{n,n})$.

Proof. 

Lemmas 1 and 2 stated that $M(G_1)=D_n^{\prime }+\delta Q_L$ and $M(G_2)=I_n+\delta Q_U$, respectively. Since minors are multiplicative,

$$M(\tilde{G})=D_n^{\prime }+D_n^{\prime }\delta Q_U+\delta Q_L+{\delta }^2{Q}_L{Q}_U.$$

Note that

$$\begin{array}{ccc}\hfill D_n^{\prime }\delta Q_U& =\left[\begin{array}{ccccc}0& q_{1,2}^U\delta & \cdots & q_{1,n-1}^U\delta & q_{1,n}^U\delta \\ 0& 0& \cdots & q_{2,n-1}^U{\delta }^2& q_{2,n}^U{\delta }^2\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 0& q_{n-1,n}^U{\delta }^2\\ 0& 0& \cdots & 0& 0\end{array}\right]\hfill & \hfill =\delta \left[\begin{array}{ccccc}0& w_{1,2}^U& \cdots & w_{1,n-1}^U& w_{1,n}^U\\ 0& 0& \cdots & w_{2,n-1}^U& w_{2,n}^U\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & 0& w_{n-1,n}^U\\ 0& 0& \cdots & 0& 0\end{array}\right]\end{array}$$

where $w_{i,j}^U=q_{i,j}^U{\delta }^k$ for $k=0$ or $k=1$. Since $q_{i,j}^U,\delta \in \mathbb{Z}$, then $w_{i,j}^U\in \mathbb{Z}$ for all $i<j$ and $i,j=1,...,n$. Thus, $D_n^{\prime }\delta Q_U=\delta W_U$, where $W_U\in {\mathbb{Z}}^{n\times n}$ with entries $w_{i,j}^U\in \mathbb{Z}$. Furthermore,

$$\begin{array}{cc}\hfill {\delta }^2{Q}_L{Q}_U& =\left[\begin{array}{ccccc}0& 0& \cdots & 0& 0\\ 0& {\delta }^2{s}_{2,2}& \cdots & {\delta }^2{s}_{2,n-1}& {\delta }^2{s}_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& {\delta }^2{s}_{n-1,2}& \cdots & {\delta }^2{s}_{n-1,n-1}& {\delta }^2{s}_{n-1,n}\\ 0& {\delta }^2{s}_{n,2}& \cdots & {\delta }^2{s}_{n,n-1}& {\delta }^2{s}_{n,n}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ccccc}0& 0& \cdots & 0& 0\\ 0& \delta s_{2,2}^{\prime }& \cdots & \delta s_{2,n-1}^{\prime }& \delta s_{2,n}^{\prime }\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& \delta s_{n-1,2}^{\prime }& \cdots & \delta s_{n-1,n-1}^{\prime }& \delta s_{n-1,n}^{\prime }\\ 0& \delta s_{n,2}^{\prime }& \cdots & \delta s_{n,n-1}^{\prime }& \delta s_{n,n}^{\prime }\end{array}\right]\hfill \end{array}$$

where $s_{i,j}^{\prime }=\delta (s_{i,j})\in \mathbb{Z}$ for all $i,j=1,\dots ,n$. Thus,

$$\begin{array}{c}\hfill D_n^{\prime }+{\delta }^2{Q}_L{Q}_U=\left[\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& \delta +\delta s_{2,2}^{\prime }& \cdots & \delta s_{2,n-1}^{\prime }& \delta s_{2,n}^{\prime }\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& \delta s_{n-1,2}^{\prime }& \cdots & \delta +\delta s_{n-1,n-1}^{\prime }& \delta s_{n-1,n}^{\prime }\\ 0& \delta s_{n,2}^{\prime }& \cdots & \delta s_{n,n-1}^{\prime }& \delta +\delta s_{n,n}^{\prime }\end{array}\right]\end{array}$$

Hence, the matrix $M(\tilde{G})$ has the following entries:

$$\begin{array}{cc}\hfill M(\tilde{G})& =(D_n^{\prime }+{\delta }^2{Q}_L{Q}_U)+\delta (Q_L+W_U)\hfill \\ \hfill & =\left[\begin{array}{ccccc}1& \delta y_{1,2}& \cdots & \delta y_{1,n-1}& \delta y_{1,n}\\ \delta y_{2,1}& \delta (1+s_{2,2}^{\prime })& \cdots & \delta y_{2,n-1}& \delta y_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \delta y_{n-1,1}& \delta y_{n-1,2}& \cdots & \delta (1+s_{n-1,n-1}^{\prime })& \delta y_{n-1,n}\\ \delta y_{n,1}& \delta y_{n,2}& \cdots & \delta y_{n,n-1}& \delta (1+s_{n,n}^{\prime })\end{array}\right]\hfill \end{array}$$

where $y_{i,j}\in \mathbb{Z}$ for all $i,j=1,\cdots ,n$. Therefore,

$$\begin{array}{cc}\hfill M(\tilde{G})& =\left[\begin{array}{ccccc}1& \delta y_{1,2}& \cdots & \delta y_{1,n-1}& \delta y_{1,n}\\ \delta y_{2,1}& \delta y_{2,2}& \cdots & \delta y_{2,n-1}& \delta y_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ \delta y_{n-1,1}& \delta y_{n-1,2}& \cdots & \delta y_{n-1,n-1}& \delta y_{n-1,n}\\ \delta y_{n,1}& \delta y_{n,2}& \cdots & \delta y_{n,n-1}& \delta y_{n,n}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& \delta y_{2,2}& \cdots & 0& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& \cdots & \delta y_{n-1,n-1}& 0\\ 0& 0& \cdots & 0& \delta y_{n,n}\end{array}\right]+\delta \left[\begin{array}{ccccc}0& y_{1,2}& \cdots & y_{1,n-1}& y_{1,n}\\ y_{2,1}& 0& \cdots & y_{2,n-1}& y_{2,n}\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ y_{n-1,1}& y_{n-1,2}& \cdots & 0& y_{n-1,n}\\ y_{n,1}& y_{n,2}& \cdots & y_{n,n-1}& 0\end{array}\right]\hfill \\ \hfill & =D_n+\delta Y_{LU}.\hfill \end{array}$$

Therefore, $C(\tilde{G})={(-1)}^{i+j}M(\tilde{G})$, where $M(\tilde{G})=D_n+\delta Y_{LU}$. □

Note that the cofactor matrix $C(\tilde{G})$ consists of integer entries with a common factor $\delta$, except for the entry $c_{1,1}(\tilde{G})=1$.

4. Inversion of the Almost-Unimodular Matrix

This section provides the proof that the inverse of the almost-unimodular matrix $\tilde{G}$, denoted as ${\tilde{G}}^{-1}\in {\mathbb{R}}^{n\times n}$, contains only a single non-integer entry. Consider the following theorem:

Theorem 3.

For $\delta \in \mathbb{Z}$ with $\delta \ne 0,\pm 1$, let $\tilde{G}\in {\mathbb{Z}}^{n\times n}$ be an almost-unimodular matrix. Then, the inverse matrix $\tilde{G}$ is ${\tilde{G}}^{-1}=D_n+{(-1)}^{i+j}{(Y_{LU})}^T$, where $D_n=diag(\frac{1}{\delta },y_{2,2},\cdots ,y_{n,n})$.

Proof. 

From Theorem 2, we have $Adj(\tilde{G})=C{(\tilde{G})}^T$, and from Lemma 3, we have $det(\tilde{G})=\delta$. Therefore,

$${\tilde{G}}^{-1}=\frac{1}{det(\tilde{G})}Adj(\tilde{G})=D_n+{(-1)}^{i+j}{(Y_{LU})}^T$$

where $D_n=diag(\frac{1}{\delta },y_{2,2},\cdots ,y_{n,n})$. □

Example 3.

For $n=5$ and $\delta =3$, let

$$G_1=\left[\begin{array}{ccccc}\delta & \delta (3)& \delta (5)& \delta (1)& \delta (-1)\\ 0& 1& \delta (2)& \delta (-5)& \delta (-2)\\ 0& 0& 1& \delta (1)& \delta (6)\\ 0& 0& 0& 1& \delta (-4)\\ 0& 0& 0& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}3& 9& 15& 3& -3\\ 0& 1& 6& -15& -6\\ 0& 0& 1& 3& 18\\ 0& 0& 0& 1& -12\\ 0& 0& 0& 0& 1\end{array}\right]$$

and

$$G_2=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ \delta (1)& 1& 0& 0& 0\\ \delta (-2)& \delta (-2)& 1& 0& 0\\ \delta (1)& \delta (3)& \delta (1)& 1& 0\\ \delta (-2)& \delta (2)& \delta (-1)& \delta (3)& 1\end{array}\right]=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 3& 1& 0& 0& 0\\ -6& -6& 1& 0& 0\\ 3& 9& 3& 1& 0\\ -6& 6& -3& 9& 1\end{array}\right].$$

Then, the almost-unimodular matrix $\tilde{G}\in {\mathbb{Z}}^{5\times 5}$ is as follows:

$$\tilde{G}=G_1{G}_2=\left[\begin{array}{ccccc}-33& -72& 33& -24& -3\\ -42& -206& -21& -69& -6\\ -105& 129& -44& 165& 18\\ 75& -63& 39& -107& -12\\ -6& 6& -3& 9& 1\end{array}\right].$$

The inverse of the almost-unimodular matrix $\tilde{G}$ is

$${\tilde{G}}^{-1}=\left[\begin{array}{ccccc}\frac{1}{3}& -3& 13& -85& -1271\\ -1& 10& -45& 288& 4323\\ -4& 42& -191& 1215& 18258\\ 20& -207& 939& -5981& -89856\\ -184& 1911& -8676& 55236& 829915\end{array}\right].$$

On the other hand, let an ordinary non-singular matrix $G\in {\mathbb{Z}}^{5\times 5}$ be the following:

$$G=\left[\begin{array}{ccccc}-81& -98& -76& -4& 29\\ -38& -77& -72& 27& 44\\ -18& 57& -2& 8& 92\\ 87& 27& -32& 69& -31\\ 33& -93& -74& 99& 67\end{array}\right].$$

The inverse of the matrix G with entries rounded to four significant figures places is

$$G^{-1}=\left[\begin{array}{ccccc}-0.1690& 0.2454& -0.03389& -0.02548& -0.05325\\ 0.03928& -0.05336& 0.01434& 0.01434& 0.004986\\ 0.07780& -0.1368& 0.01258& 0.0004842& 0.03909\\ 0.2009& -0.3049& 0.0394& 0.03369& 0.07471\\ -0.07318& 0.1046& -0.007802& -0.01679& -0.01916\end{array}\right].$$

Without being rounded, the matrix $G^{-1}$ has longer entries. For instance, let $g_{i,j}^{\prime }$ denote the entries of $G^{-1}$ for $i,j=1,\cdots ,5$. Then, a few entries of the matrix $G^{-1}$ that are rounded to 30 significant figures are given as follows:

$$\begin{array}{cc}\hfill g_{1,1}^{\prime }& =-0.168993685053825752126666924276,\hfill \\ \hfill g_{2,1}^{\prime }& =0.0392803016804847539890289920868,\hfill \\ \hfill g_{3,1}^{\prime }& =0.0778045107903581037237408795958,\hfill \\ \hfill g_{5,5}^{\prime }& =-0.0191552177357843254835474649937.\hfill \end{array}$$

Observe that all entries in the matrix $G^{-1}$ are long floating-point numbers. On the other hand, the inverse of $\tilde{G}$ has only a single non-integer entry, which is ${\tilde{g}}_{1,1}^{\prime }=\frac{1}{3}$. Clearly, there is a significant difference between the inverse of the integer matrix G and the inverse of the almost-unimodular matrix $\tilde{G}$.

5. Conclusions

In this paper, we proposed a matrix $\tilde{G}\in {\mathbb{Z}}^{n\times n}$—referred to as the almost-unimodular matrix—with inverse ${\tilde{G}}^{-1}\in {\mathbb{R}}^{n\times n}$ that was proven to contain only a single non-integer entry, which is the first entry ${\tilde{g}}_{1,1}^{\prime }\in \mathbb{R}$. For large values of n, this property could be a significant advantage, since the matrix inversion operation can be performed with minimal involvement of non-integer entries. With only a single non-integer entry, storing or transmitting the inverse matrix ${\tilde{G}}^{-1}\in {\mathbb{R}}^{n\times n}$ might be easier than before. Further investigation of the true potential of the matrix $\tilde{G}$ could be explored. The computational cost of the inversion operation involving $\tilde{G}$, as well as the size of the inverse matrix ${\tilde{G}}^{-1}$, should be determined for various sizes of n. Other than that, the application of the matrix $\tilde{G}$ in any area where the inversion of a large integer matrix is necessary could also be a research opportunity with great potential in the future.