基本的な行列の問題

■問題

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

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

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

■解答

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

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

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

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

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

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

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

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

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

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

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

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

(2)より

${\displaystyle Q}$${\displaystyle =P-3B}$…(3)

(3)を(1)に代入する．

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

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

${\displaystyle P}$${\displaystyle =\frac{1}{4}A+\frac{3}{2}B}$…(4)

(4)を(3)に代入する

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

${\displaystyle Q}$${\displaystyle =\frac{1}{4}A-\frac{3}{2}B}$…(5)

(4)，(5)に${\displaystyle A}$ と${\displaystyle B}$ の具体的な行列を代入する．

${\displaystyle P}$${\displaystyle =\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)}$

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

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

${\displaystyle Q}$${\displaystyle =\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)}$

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

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

 

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作成：学生スタッフ

最終更新日： 2022年9月7日