因此，我写了一些解决方案，但行列式却从bo中流出。如何解决这个问题？
我正在使用 Dyan Yu 的 Sty 文件，可以通过快速谷歌搜索在 github 上找到它。
\documentclass[paper=6.125in:8.25in,twoside,openany,pagesize=pdftex,10pt]{scrbook}
\usepackage[retro,cabin,sleek,nosecthm]{dylanadi}
\usepackage[top=1cm, bottom=1.5cm, left=1cm, right=1cm]{geometry}
\usepackage{enumitem}
\usepackage{lmodern}
\title{KVPY SA Mathematics 2022}
\author{Ayan Khan}
\date{\today}
\date{\today}
\changemaincolor{blue}
\changesecondcolor{red}
\begin{document}
\begingroup % figure out how to put logo in
\notofont
\color{\secondcolor}
\let\newpage\relax%
\dylantitle
\endgroup
\newpage
%\tableofcontents
\newpage
\begin{problem}
\newline \newline
Let A be a matrix defined as $A = \displaystyle \begin{bmatrix}
a_{1} & a_{2} & a_{3} & \dots & a_{n} \\
a_{n} & a_{1} & x_{2} & \dots & a_{n-1} \\
a_{n-1} & a_{n} & x_{1} & \dots & a_{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{2} & a_{3} & x_{4} & \dots & a_{1}
\end{bmatrix}$ If $g(x)$ is equal to $\displaystyle a_{1}+a_{2}x+a_{3}x^{2}+\cdots+a_{n}x^{n-1}$ and $a_{i}$, where i = 1 to n, denote the $n^{th}$ complex root of unity. If $A ={\displaystyle \prod_{i=1}^{n} g(\alpha_{i})}$ and $b_{0}=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and $\displaystyle b_{n} = \frac{b_{n}^{2}-5}{2(b_{n}+2)}$, $\displaystyle \forall n > 0,$ then porve that $\displaystyle b_{n} = cot\Big(\frac{2^{n-3}\pi}{3}\Big)-2k$.
\end{problem}
\begin{solution}
As the coefficients of $g(x)$ from the Determinant $A$, we shall solve this by assuming an another Determinant $\displaystyle Y =\displaystyle \begin{vmatrix}
1 & \alpha_{1} & \alpha_{1}^{2} & \dots & \alpha_{1}^{n-1} \\
1 & \alpha_{2} & \alpha_{2}^{2} & \dots & \alpha_{2}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{n} & \alpha_{n}^{2} & \dots & \alpha_{n}^{n-1}
\end{vmatrix}$
\newline Now,
$A \times Y =\begin{vmatrix}
a_{1}+a_{2}\alpha_{1}+a_{3}\alpha_{1}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} & a_{1}+a_{2}\alpha_{2}+a_{n}\alpha_{2}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} & \alpha_{1}^{2} & \dots & \alpha_{1}^{n-1} \\
1 & \alpha_{2} & \alpha_{2}^{2} & \dots & \alpha_{2}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{n} & \alpha_{n}^{2} & \dots & \alpha_{n}^{n-1}
\end{vmatrix}$
\end{solution}
\end{document}
答案1
我将重点介绍最后的代码块，即
\newline Now,
$A \times Y =\begin{vmatrix}
a_{1}+a_{2}\alpha_{1}+a_{3}\alpha_{1}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} & a_{1}+a_{2}\alpha_{2}+a_{n}\alpha_{2}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} & \alpha_{1}^{2} & \dots & \alpha_{1}^{n-1} \\
1 & \alpha_{2} & \alpha_{2}^{2} & \dots & \alpha_{2}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{n} & \alpha_{n}^{2} & \dots & \alpha_{n}^{n-1}
\end{vmatrix}$
唯一可行的排版方法是使用矩阵 (1,1) 和 (1,2) 位置元素的快捷方式，然后再拼出快捷方式。
请养成使用真实显示数学模式而不是$\displaystyle...$构造的习惯。
以下屏幕截图中的框线之所以存在，是因为该geometry包中加载了选项showframe。
（题外话：我认为我可以发现最后一个等式中的两个错误。）
\documentclass[paper=6.125in:8.25in,10pt]{scrbook}
%%\usepackage[retro,cabin,sleek,nosecthm]{dylanadi}
\usepackage{newpxtext,newpxmath}
\usepackage{mathtools}
\usepackage[top=1cm, bottom=1.5cm, hmargin=1cm, showframe]{geometry}
\usepackage{xcolor}
\begin{document}
\noindent
\textcolor{red}{\Large Before}
Now,
$A \times Y =
\begin{vmatrix}
a_{1}+a_{2}\alpha_{1}+a_{3}\alpha_{1}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} &
a_{1}+a_{2}\alpha_{2}+a_{n}\alpha_{2}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} &
\alpha_{1}^{2} & \dots & \alpha_{1}^{n-1} \\
1 & \alpha_{2} & \alpha_{2}^{2} & \dots & \alpha_{2}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{n} & \alpha_{n}^{2} & \dots & \alpha_{n}^{n-1}
\end{vmatrix}$
\bigskip\noindent
\textcolor{red}{\Large After}
\medskip
Now,
\[
\det K\equiv \det (A \times Y) =\begin{vmatrix}
\kappa_{11} & \kappa_{12} & \alpha_{1}^{2} & \dots & \alpha_{1}^{n-1} \\
1 & \alpha_{2} & \alpha_{2}^{2} & \dots & \alpha_{2}^{n-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_{n} & \alpha_{n}^{2} & \dots & \alpha_{n}^{n-1}
\end{vmatrix}
\]
where
\begin{align*}
\kappa_{11}&=a_{1}+a_{2}\alpha_{1}+a_{3}\alpha_{1}^{2}+\cdots+a_{n}\alpha_{1}^{n-1} \\
\shortintertext{and}
\kappa_{12}&=a_{1}+a_{2}\alpha_{2}+a_{n}\alpha_{2}^{2}+\cdots+a_{n}\alpha_{1}^{n-1}\,.
\end{align*}
\end{document}