# 表現行列に関する問題

## ■問題

•  $\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)\to \left(\begin{array}{c}{x}_{1}+2{x}_{2}\\ {x}_{2}-{x}_{3}\end{array}\right)$
•  $g：{R}^{2}\to {R}^{2}$ $\left(\begin{array}{c}{{x}_{1}}^{\prime }\\ {{x}_{2}}^{\prime }\end{array}\right)\to \left(\begin{array}{c}{{x}_{1}}^{\prime }-{{x}_{2}}^{\prime }\\ -2{{x}_{2}}^{\prime }\end{array}\right)$

$f$$g$表現行列 $A$$B$を求めよ．

## ■答

$A=\left(\begin{array}{c}\begin{array}{ccc}1& 2& 0\end{array}\\ \begin{array}{ccc}0& 1& -1\end{array}\end{array}\right)$$B=\left(\begin{array}{cc}1& -1\\ 0& -2\end{array}\right)$

## ■計算

$\left(\begin{array}{c}{x}_{1}+2{x}_{2}\\ {x}_{2}-{x}_{3}\end{array}\right)$$=\left(\begin{array}{c}\begin{array}{ccc}1& 2& 0\end{array}\\ \begin{array}{ccc}0& 1& -1\end{array}\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)$　･･････(1)

よって，$f$の表現行列$A$

$A=\left(\begin{array}{c}\begin{array}{ccc}1& 2& 0\end{array}\\ \begin{array}{ccc}0& 1& -1\end{array}\end{array}\right)$

となる．

$\left(\begin{array}{c}{{x}_{1}}^{\prime }-{{x}_{2}}^{\prime }\\ -2{{x}_{2}}^{\prime }\end{array}\right)$ $=\left(\begin{array}{cc}1& -1\\ 0& -2\end{array}\right)\left(\begin{array}{c}{{x}_{1}}^{\prime }\\ {{x}_{2}}^{\prime }\end{array}\right)$　･･････(2)

よって，$g$の表現行列$B$

$B=\left(\begin{array}{cc}1& -1\\ 0& -2\end{array}\right)$

となる．

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