# 基本的な行列の問題

## ■問題

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

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

[ii]2つの等式$2\left(P+Q\right)=A$$\frac{1}{3}\left(P-Q\right)=B$を同時に満たす2次の正方行列$P$$Q$

## ■解答

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

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

${2A}^{2}-6X$$=\left\{2{B}^{2}-4\left(X-2{A}^{2}\right)\right\}$

$4X-6X$$={2{B}^{2}+8{A}^{2}-2A}^{2}$

$-\mathrm{2X}$$=6{A}^{2}+2{B}^{2}$

$X$$=-3{A}^{2}-{B}^{2}$

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

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

$=-3\left(\begin{array}{cc}19& -10\\ -4& 11\end{array}\right)-\left(\begin{array}{cc}7& -27\\ -9& 52\end{array}\right)$

$=\left(\begin{array}{cc}-57& 30\\ 12& -33\end{array}\right)-\left(\begin{array}{cc}7& -27\\ -9& 52\end{array}\right)$

$=\left(\begin{array}{cc}-64& 57\\ 21& -85\end{array}\right)$

[ii]2つの等式$2\left(P+Q\right)=A$ …(1) ，$\frac{1}{3}\left(P-Q\right)=B$ …(2) を同時に満たす2次の正方行列$P$$Q$ を求めよ．

(2)より

$Q$$=P-3B$…(3)

(3)を(1)に代入する．

$2\left\{P+\left(P-3B\right)\right\}=A$

$4P$$=A+\mathrm{6B}$

$P$$=\frac{1}{4}A+\frac{3}{2}B$…(4)

(4)を(3)に代入する

$Q=\frac{1}{4}A+\frac{3}{2}B-3B$

$Q$$=\frac{1}{4}A-\frac{3}{2}B$…(5)

(4)，(5)に$A$$B$ の具体的な行列を代入する．

$P$$=\frac{1}{4}\left(\begin{array}{cc}-3& 5\\ 2& 1\end{array}\right)+\frac{3}{2}\left(\begin{array}{cc}2& -3\\ -1& 7\end{array}\right)$

$=\left(\begin{array}{cc}-\frac{3}{4}& \frac{5}{4}\\ \frac{1}{2}& \frac{1}{4}\end{array}\right)+\left(\begin{array}{cc}3& -\frac{9}{2}\\ -\frac{3}{2}& \frac{21}{2}\end{array}\right)$

$=\left(\begin{array}{cc}\frac{9}{4}& -\frac{13}{4}\\ -1& \frac{43}{4}\end{array}\right)$

$Q$$=\frac{1}{4}\left(\begin{array}{cc}-3& 5\\ 2& 1\end{array}\right)-\frac{3}{2}\left(\begin{array}{cc}2& -3\\ -1& 7\end{array}\right)$

$=\left(\begin{array}{cc}-\frac{3}{4}& \frac{5}{4}\\ \frac{1}{2}& \frac{1}{4}\end{array}\right)-\left(\begin{array}{cc}3& -\frac{9}{2}\\ -\frac{3}{2}& \frac{21}{2}\end{array}\right)$

$=\left(\begin{array}{cc}-\frac{15}{4}& \frac{23}{4}\\ 2& -\frac{41}{4}\end{array}\right)$

ホーム>>カテゴリー分類>>行列>>線形代数>>線形代数に関する問題>>行列の計算>>問題