类似三对角矩阵行列式的递推公式算法

doi:10.12677/PM.2022.1211200

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类似三对角矩阵行列式的递推公式算法
A Recursive Formula Algorithm of Similar to a Tridiagonal Matrix Determinant

DOI: 10.12677/PM.2022.1211200, PDF, HTML, XML, 下载: 153 浏览: 834 科研立项经费支持
作者: 左陈宇, 代慧慧, 李梦玟, 张雨婷, 袁之成, 董兵, 王文：合肥师范学院数学与统计学院，安徽合肥
关键词: 三对角行列式；迭代；递推；Tridiagonal Determinant； Iteration； Recursion

摘要: 本文研究了一类特殊的类似三对角的行列式的递推公式。通过迭代关系算法表示，得到一个简洁的递推结果表示，方便在其它领域的应用。

Abstract: In this paper, we study the recurrence formula of a special kind of determinant similar to tridiagonal. A concise recursive result representation is obtained by using iterative relation algo-rithm, which is convenient for application in other fields.

文章引用：左陈宇, 代慧慧, 李梦玟, 张雨婷, 袁之成, 董兵, 王文. 类似三对角矩阵行列式的递推公式算法[J]. 理论数学, 2022, 12(11): 1869-1874. https://doi.org/10.12677/PM.2022.1211200

1. 引言及主要结果

行列式的计算是高等代数中十分重要的内容 [1] ，而且一些特殊的行列式在实际应用中有着重要的作用。比如三对角行列式在经济学中有着非常广泛的应用，在线性代数和组合数学中也具有重要的应用价值。因此，寻找这些特殊行列式的显示计算公式或地推公式也变得非常重要。文献 [2] 中介绍一些三阶的三对角矩阵的行列式的计算方法，文献 [3] [4] 中介绍对角矩阵逆矩阵的计算公式。文献 [5] 中给出n阶三对角矩阵的行列式特殊形式的计算公式，文献 [3] 中给出了下面一般的类似n阶三对角矩阵U的行列式的递推公式和连分数形式的公式，

$U = (\begin{matrix} a_{1} & c_{1} & 0 & 0 & \dots & 0 & 0 & 0 & d_{n - 2} \\ a_{2} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 & d_{n - 3} \\ a_{3} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 & d_{n - 4} \\ a_{4} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 & d_{n - 5} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ & ⋮ \\ a_{n - 2} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} & d_{1} \\ a_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} & c_{n - 1} \\ a_{n} & 0 & 0 & 0 & \dots & 0 & 0 & 0 & b_{n} \end{matrix})$

本文主要受文献 [3] 的启发，对矩阵U再加一条边，并给出具体的递推计算，并对其进行简洁表示。本文构造的行列式有三条边框，即第一行、第一列和第 $n + 1$ 列，其中的元素是互异的常数。另一条边框，即第 $n + 1$ 行，其第一个和最后一个元素为常数，且与行列式中各元素均互异，第 $n + 1$ 行的其它元素均为0，即

$D = (\begin{matrix} a_{1} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} & d_{n - 2} \\ a_{2} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 & d_{n - 3} \\ a_{3} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 & d_{n - 4} \\ a_{4} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 & d_{n - 5} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ & ⋮ \\ a_{n - 2} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} & d_{1} \\ a_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} & c_{n - 1} \\ a_{n} & 0 & 0 & 0 & \dots & 0 & 0 & 0 & b_{n} \end{matrix})$

进一步地，所构造行列式的计算结果存在叠代关系。最终发现此关系可用秦九韶算法表示，即得出简洁的结果表示，方便在其它领域的应用。我们的主要结果如下：

定理对 $n \in N^{+}$ ，矩阵D的行列式可表示为

$| D_{n} | = - a_{n} \cdot (P_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{P_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n - 1} b_{i} + (Q_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{Q_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n} b_{i},$ (1)

其中

$\begin{array}{l} P_{1} = c_{n - 1}, P_{i} = - \frac{c_{(n - i)}}{b_{(n - i + 1)}} \cdot P_{i - 1} + d_{i - 1}; \\ Q_{1} = a_{n - 1}, Q_{i} = - \frac{c_{(n - i)}}{b_{(n - i + 1)}} \cdot Q_{i - 1} + a_{n - i}, i = 2, 3, \dots, n - 1. \end{array}$

证明：按最后一行展开可得

$\begin{matrix} | D | = {(- 1)}^{n + 1} a_{n} | \begin{matrix} c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} & d_{n - 2} \\ b_{2} & c_{2} & c_{2} & \dots & 0 & 0 & 0 & d_{n - 3} \\ 0 & b_{3} & b_{3} & \dots & 0 & 0 & 0 & d_{n - 4} \\ 0 & 0 & b_{4} & \dots & 0 & 0 & 0 & d_{n - 5} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} & d_{1} \\ 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} & c_{n - 1} \end{matrix} | + b_{n} | \begin{matrix} a_{1} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ a_{2} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ a_{3} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ a_{4} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ a_{n - 2} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} \\ a_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} | \\ = H_{1} + H_{2} \end{matrix}$

下面分别计算行列式 $H_{1}$ 和 $H_{2}$ 。先计算 $H_{1}$ ，添加边框后再交换第一列和最后一列，可得

$H_{1} = | \begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} 0 \\ c_{1} \end{array} & \begin{array}{l} 0 \\ e_{1} \end{array} & \begin{array}{l} 0 \\ e_{2} \end{array} & \begin{array}{l} \dots \\ \dots \end{array} & \begin{array}{l} 0 \\ e_{n - 4} \end{array} & \begin{array}{l} 0 \\ e_{n - 3} \end{array} & \begin{array}{l} 0 \\ d_{n - 2} \end{array} \\ 0 & b_{2} & c_{2} & 0 & \dots & 0 & 0 & d_{n - 3} \\ 0 & 0 & b_{3} & c_{3} & \dots & 0 & 0 & d_{n - 4} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & \dots & c_{n - 3} & 0 & d_{2} \\ 0 & 0 & 0 & 0 & \dots & b_{n - 2} & c_{n - 2} & d_{1} \\ 0 & 0 & 0 & 0 & \dots & 0 & b_{n - 1} & c_{n - 1} \end{matrix} | = - | \begin{matrix} \begin{array}{l} 0 \\ d_{n - 2} \end{array} & \begin{array}{l} 0 \\ c_{1} \end{array} & \begin{array}{l} 0 \\ e_{1} \end{array} & \begin{array}{l} 0 \\ e_{2} \end{array} & \begin{array}{l} \dots \\ \dots \end{array} & \begin{array}{l} 0 \\ e_{n - 4} \end{array} & \begin{array}{l} 0 \\ e_{n - 3} \end{array} & \begin{array}{l} 1 \\ 0 \end{array} \\ d_{n - 3} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ d_{n - 4} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ d_{2} & 0 & 0 & 0 & \dots & c_{n - 3} & 0 & 0 \\ d_{1} & 0 & 0 & 0 & \dots & b_{n - 2} & c_{n - 2} & 0 \\ c_{n - 1} & 0 & 0 & 0 & \dots & 0 & b_{n - 1} & 0 \end{matrix} |$

$\underline{\underline{按第一行展开}} {(- 1)}^{n + 2} | \begin{matrix} d_{n - 2} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ d_{n - 3} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ d_{n - 4} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ d_{n - 5} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ d_{1} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} \\ c_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |$

$= {(- 1)}^{n + 2} | \begin{matrix} d_{n - 2} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ d_{n - 3} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ d_{n - 4} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ d_{n - 5} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ - \frac{c_{n - 2}}{b_{n - 2}} c_{n - 1} + d_{1} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & 0 \\ c_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |$

$= {(- 1)}^{n + 2} | \begin{matrix} d_{n - 2} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ d_{n - 3} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ d_{n - 4} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ - (- \frac{c_{n - 2}}{b_{n - 2}} c_{n - 1} + d_{1}) \frac{c_{n - 3}}{b_{n - 2}} + d_{2} & 0 & 0 & 0 & \dots & b_{n - 3} & 0 & 0 \\ - \frac{c_{n - 2}}{b_{n - 2}} c_{n - 1} + d_{1} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & 0 \\ c_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |$

令 $P_{1} = c_{n - 1}, P_{2} = - \frac{c_{n - 2}}{b_{n - 1}} c_{n - 1} + d_{1}, P_{3} = - \frac{c_{n - 3}}{b_{n - 2}} (- \frac{c_{n - 2}}{b_{n - 1}} c_{n - 1} + d_{1}) + d_{2}, \dots$

由此通项公式可以表示为

$P_{i} = - \frac{c_{n - i}}{b_{n - (i - 1)}} P_{i - 1} + d_{i - 1}, i = 2, 3, \dots, n - 1.$

则上式可化为

$H_{1} = | \begin{matrix} P_{n - 1} & 0 & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ P_{n - 2} & b_{2} & 0 & 0 & \dots & 0 & 0 & 0 \\ P_{n - 3} & 0 & b_{3} & 0 & \dots & 0 & 0 & 0 \\ P_{n - 4} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ P_{3} & 0 & 0 & 0 & \dots & b_{n - 3} & 0 & 0 \\ P_{2} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & 0 \\ P_{1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} | \underline{\underline{将第 i 列分别乘上 (- \frac{P_{n - i}}{b_{i}}) 加到第一列}}$

$| \begin{matrix} P_{n - 1} - \frac{P_{n - 3}}{b_{3}} e_{1} - \frac{P_{n - 4}}{b_{4}} e_{2} - \dots - \frac{P_{1}}{b_{n - 1}} e_{n - 3} & 0 & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ 0 & b_{2} & 0 & 0 & \dots & 0 & 0 & 0 \\ 0 & 0 & b_{3} & 0 & \dots & 0 & 0 & 0 \\ 0 & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & \dots & b_{n - 3} & 0 & 0 \\ 0 & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & 0 \\ 0 & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |$

$= (P_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{P_{n - (i + 2)}}{b_{i + 2}}) \prod_{i = 2}^{n - i} b_{i} .$

由行列式 $H_{1}$ 的计算过程中发现，下面行列式可表示为

${| \begin{matrix} d_{n - 2} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ d_{n - 3} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ d_{n - 4} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ d_{n - 5} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ d_{1} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} \\ c_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |}_{(n - 1) \times (n - 1)} = (P_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{P_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n - 1} b_{i};$ (2)

利用(2)式，可求得

$H_{2} = {| \begin{matrix} a_{1} & c_{1} & e_{1} & e_{2} & \dots & e_{n - 5} & e_{n - 4} & e_{n - 3} \\ a_{2} & b_{2} & c_{2} & 0 & \dots & 0 & 0 & 0 \\ a_{3} & 0 & b_{3} & c_{3} & \dots & 0 & 0 & 0 \\ a_{4} & 0 & 0 & b_{4} & \dots & 0 & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ \\ a_{n - 2} & 0 & 0 & 0 & \dots & 0 & b_{n - 2} & c_{n - 2} \\ a_{n - 1} & 0 & 0 & 0 & \dots & 0 & 0 & b_{n - 1} \end{matrix} |}_{(n - 1) \times (n - 1)} = (Q_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{Q_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n - 1} b_{i};$

其中 $Q_{1} = a_{n - 1}, Q_{i} = - \frac{c_{(n - i)}}{b_{(n - i + 1)}} \cdot Q_{i - 1} + a_{n - i}, i = 2, 3, \dots, n - 1$ .

所以

$\begin{matrix} | D_{n} | = {(- 1)}^{n - 1} a_{n} \cdot {(- 1)}^{n + 2} | H_{1} | + {(- 1)}^{2 n} b_{n} \cdot | H_{2} | \\ = - a_{n} \cdot (P_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{P_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n - 1} b_{i} + (Q_{n - 1} - \sum_{i = 1}^{n - 3} e_{i} \cdot \frac{Q_{[n - (i + 2)]}}{b_{i + 2}}) \cdot \prod_{i = 2}^{n} b_{i} . \end{matrix}$

2. 简单应用

例：计算行列式

$| T | = | \begin{matrix} a & β & e & e & \dots & e & e & e & d \\ a & α & β & 0 & \dots & 0 & 0 & 0 & d \\ a & 0 & α & β & \dots & 0 & 0 & 0 & d \\ a & 0 & 0 & α & \dots & 0 & 0 & 0 & d \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ & ⋮ & ⋮ \\ a & 0 & 0 & 0 & \dots & 0 & α & β & d \\ a & 0 & 0 & 0 & \dots & 0 & 0 & α & β \\ a & 0 & 0 & 0 & \dots & 0 & 0 & 0 & α \end{matrix} |$

解：根据定理中的公式可得

$\begin{matrix} | T | = - a \cdot (A_{n - 1} - \sum_{i = 1}^{n - 3} e \cdot \frac{A_{[n - (i + 2)]}}{α}) \cdot α^{n - 3} + (B_{n - 1} - \sum_{i = 1}^{n - 3} e \cdot \frac{B_{[n - (i + 2)]}}{α}) \cdot α^{n - 2} \\ = - a \cdot (A_{n - 1} + \frac{e}{α} \sum_{i = 1}^{n - 3} A_{[n - (i + 2)]}) \cdot α^{n - 3} + (B_{n - 1} - \frac{e}{α} \sum_{i = 1}^{n - 3} B_{[n - (i + 2)]}) \cdot α^{n - 2} \end{matrix}$

(其中 $A_{1} = β$ , $A_{i} = - \frac{β}{α} \cdot A_{i - 1} + d$ ; $B_{1} = a$ , $B_{i} = - \frac{β}{α} \cdot B_{i - 1} + a$ , $i = 2, 3, \dots, n - 1$ )。

基金项目

安徽省2020年创新创业训练计划省级项目(S202014098010)，2021年国家级大学生创新创业训练计划项目(202114098036)。

参考文献

[1]	北京大学数学系. 高等代数[M]. 北京: 高等教育出版社, 2019.
[2]	杨胜良. 三对角行列式及其应用[J]. 工科数学, 2002(2): 106-108.
[3]	Kılıç, E. (2008) Explicit Formula for the Inverse of a Tridiagonal Matrix by Backward Continued Fractions. Applied Mathematics and Computation, 197, 345-357. https://doi.org/10.1016/j.amc.2007.07.046
[4]	Barrett, W.W. (1979) A Theorem on Inverses of Tridiagonal Ma-trices. Linear Algebra and Its Applications, 27, 211-217. https://doi.org/10.1016/0024-3795(79)90043-0
[5]	Qi, F., Wang, W., Guo, B.-N. and Lim, D. (2021) Several Explicit and Recurrent Formulas for Determinants of Tridiagonal Matrices via Generalized Continued Fractions. In: Hammouch, Z., Dutta, H., Melliani, S. and Ruzhansky, M., Eds., Nonlinear Analysis: Problems, Applications and Computational Methods, Lecture Notes in Networks and Systems, Springer, Cham, Vol. 168, 233-248. https://doi.org/10.1007/978-3-030-62299-2_15

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