Re: \begin{align*}\end{align*}の前後で１行分スペースができる

名前: エプソム
日時: 2012-02-09 08:32:00
IPアドレス: 211.12.192.*

>>56639
サンプルソースを提示します。検証よろしくお願いします。

\documentclass[a4paper,10pt,fleqn]{jarticle}
\usepackage{amsmath}
\begin{document}

$A=\left( \begin{array}{cc}0 & 1 \\ 1 & 0 \end{array} \right),\ B=\left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right)$ のとき

%%%%%
$(1)$ $A^m=E,\ B^n=E$ となる最小の正の整数 $m,\ n$ をそれぞれ求めよ。
\begin{align*}
A^2 & = \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right)\left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right)=\left( \begin{array}{cc}1 & 0 \\ 0 & 1\end{array} \right) \\
B^2 & = \left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right)\left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right)=\left( \begin{array}{cc}-1 & 0 \\ 0 & -1\end{array} \right) \neq E \\
B^3 & = B^2 B = \left( \begin{array}{cc}-1 & 0 \\ 0 & -1\end{array} \right) \left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right)  = \left( \begin{array}{cc}0 & -1 \\ -1 & 0\end{array} \right)  \neq E \\
B^4 & = B^3 B = \left( \begin{array}{cc}0 & -1 \\ -1 & 0\end{array} \right) \left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right) = \left( \begin{array}{cc}1 & 0 \\ 0 & 1\end{array} \right)
\end{align*}
よって， $A^m=E,\ B^n=E$ となる最小の正の整数 $m,\ n$ は $m=2,\ n=4$

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$(2)$ \ $ABA=B^k,\ AB^2A=B^l$ をみたす最小の正の整数 $k,\ l$ をそれぞれ求めよ。
\begin{align*}
ABA & = \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \left( \begin{array}{cc}0 & 1 \\ -1 & 0\end{array} \right) \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \\
	 & = \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \left( \begin{array}{cc}1 & 0 \\ 0 & -1\end{array} \right) = \left( \begin{array}{cc}0 & -1 \\ 1 & 0\end{array} \right)=B^3 \\
AB^2A & = \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \left( \begin{array}{cc}-1 & 0 \\ 0 & -1\end{array} \right) \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \\
	 & = \left( \begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \left( \begin{array}{cc}0 & -1 \\ -1 & 0\end{array} \right) = \left( \begin{array}{cc}-1 & 0 \\ 0 & -1\end{array} \right)=B^2 \\
\end{align*}
よって， $ABA=B^k,\ AB^2A=B^l$ みたす最小の正の整数 $k,\ l$ は $k=3,\ l=2$

\end{document}

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