# 基本的な行列の問題

## ■問題

$A=\left(\begin{array}{ccc}1& 3& 2\\ 2& -1& 3\\ 3& 2& 1\end{array}\right)$

## ■答

$\left(\begin{array}{ccc}-\frac{1}{4}& \frac{1}{28}& \frac{11}{28}\\ \frac{1}{4}& -\frac{5}{28}& \frac{1}{28}\\ \frac{1}{4}& \frac{1}{4}& -\frac{1}{4}\end{array}\right)$

## ■計算

まず , $\left|A\right|$を求める．

$\left|A\right|$$=\left|\begin{array}{ccc}1& 3& 2\\ 2& -1& 3\\ 3& 2& 1\end{array}\right|$

$=\left|\begin{array}{ccc}1& 3& 2\\ 0& -7& -1\\ 0& -7& -5\end{array}\right|$

$=\left|\begin{array}{cc}-7& -1\\ -7& -5\end{array}\right|$

$=28$$\ne 0$

${\stackrel{˜}{a}}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right|=-7$${\stackrel{˜}{a}}_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 3\\ 3& 1\end{array}\right|=7$${\stackrel{˜}{a}}_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& -1\\ 3& 2\end{array}\right|=7$

${\stackrel{˜}{a}}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 2\\ 2& 1\end{array}\right|=1$${\stackrel{˜}{a}}_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 2\\ 3& 1\end{array}\right|=-5$${\stackrel{˜}{a}}_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 3\\ 3& 2\end{array}\right|=7$

${\stackrel{˜}{a}}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 2\\ -1& 3\end{array}\right|=11$${\stackrel{˜}{a}}_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 2\\ 2& 3\end{array}\right|=1$${\stackrel{˜}{a}}_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 3\\ 2& -1\end{array}\right|=-7$

${A}^{-1}=\frac{1}{|A|}\left(\begin{array}{cccc}{\stackrel{˜}{a}}_{11}& {\stackrel{˜}{a}}_{21}& \cdots & {\stackrel{˜}{a}}_{n1}\\ {\stackrel{˜}{a}}_{12}& {\stackrel{˜}{a}}_{22}& \cdots & {\stackrel{˜}{a}}_{n2}\\ ⋮& ⋮& & ⋮\\ {\stackrel{˜}{a}}_{1n}& {\stackrel{˜}{a}}_{2n}& \cdots & {\stackrel{˜}{a}}_{nn}\end{array}\right)$

に代入する.この問題では$n=3$ である．

${A}^{-1}$ $=\frac{1}{28}\left(\begin{array}{ccc}-7& 1& 11\\ 7& -5& 1\\ 7& 7& -7\end{array}\right)$

$=\left(\begin{array}{ccc}-\frac{7}{28}& \frac{1}{28}& \frac{11}{28}\\ \frac{7}{28}& -\frac{5}{28}& \frac{1}{28}\\ \frac{7}{28}& \frac{7}{28}& -\frac{7}{28}\end{array}\right)$

$=\left(\begin{array}{ccc}-\frac{1}{4}& \frac{1}{28}& \frac{11}{28}\\ \frac{1}{4}& -\frac{5}{28}& \frac{1}{28}\\ \frac{1}{4}& \frac{1}{4}& -\frac{1}{4}\end{array}\right)$

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