# 基本的な行列の問題

## ■問題

$A=\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)$$B=\left(\begin{array}{cc}-2& 5\\ 1& -9\end{array}\right)$ のとき，次の行列を求めよ．

[i]等式$\frac{2}{5}A-\frac{2}{3}X=\frac{1}{6}\left\{-2A-\left(2X-3B\right)\right\}$ を満たす行列$X$

[ii]2つの等式$PQ=A$…(1) ，$P=A+B$…(2) を同時に満たす2次の正方行列$P$$Q$を求めよ．

## ■解答

[i]等式$\frac{2}{5}A-\frac{2}{3}X=\frac{1}{6}\left\{-2A-\left(2X-3B\right)\right\}$ を満たす行列$X$

$\frac{2}{5}A-\frac{2}{3}X$$=\frac{1}{6}\left\{-2A-\left(2X-3B\right)\right\}$

$\frac{1}{30}\left(12A-20X\right)$$=\frac{5}{30}\left\{-2A-\left(2X-3B\right)\right\}$

$12A-20X$$=5\left\{-2A-\left(2X-3B\right)\right\}$

$12A-20X$$=-10A-10X+15B$

$-10X$$=-22A+15B$

$X$$=\frac{11}{5}A-\frac{3}{2}B$

$=\frac{11}{5}\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)-\frac{3}{2}\left(\begin{array}{cc}-2& 5\\ 1& -9\end{array}\right)$

$=\left(\begin{array}{cc}11& \frac{33}{5}\\ -\frac{77}{5}& -22\end{array}\right)+\left(\begin{array}{cc}3& -\frac{15}{2}\\ -\frac{3}{2}& \frac{27}{2}\end{array}\right)$

$=\left(\begin{array}{cc}14& -\frac{9}{10}\\ -\frac{169}{10}& -\frac{17}{2}\end{array}\right)$

[ii]2つの等式$PQ=A$…(1) ，$P=A+B$…(2) を同時に満たす2次の正方行列$P$$Q$を求めよ．

(1)より

$P$$=\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)+\left(\begin{array}{cc}-2& 5\\ 1& -9\end{array}\right)$$=\left(\begin{array}{cc}3& 8\\ -6& -19\end{array}\right)$

(2)より

$Q$$={P}^{-1}A$

$={\left(\begin{array}{cc}3& 8\\ -6& -19\end{array}\right)}^{-1}\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)$

$=\frac{1}{-9}\left(\begin{array}{cc}-19& -8\\ 6& 3\end{array}\right)\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)$

$=-\frac{1}{9}\left(\begin{array}{cc}-19& -8\\ 6& 3\end{array}\right)\left(\begin{array}{cc}5& 3\\ -7& -10\end{array}\right)$

$=-\frac{1}{9}\left(\begin{array}{cc}-39& 23\\ 9& -12\end{array}\right)$

$=\left(\begin{array}{cc}\frac{13}{3}& -\frac{23}{9}\\ -1& -\frac{4}{3}\end{array}\right)$

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