基本的な行列の問題

■問題

$3$\displaystyle{ A=\left({\begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right) }$ ， $3$\displaystyle{ B=\left({\begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array}}\right) }$ のとき，次の行列を求めよ．

[i]等式 $3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{A}^{2}- X=\frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}\left{{ 2{B}^{2}- 4\left({ X- 2{A}^{2} }\right) }\right} }$ を満たす行列 $3$\displaystyle{ X }$

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

■解答

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{A}^{2}- X=\frac{\hspace{2}1\hspace{2}}{\hspace{2}6\hspace{2}}\left{{ 2{B}^{2}- 4\left({ X- 2{A}^{2} }\right) }\right} }$

$3$\displaystyle{ { 2A }^{2}- 6X }$ $3$\displaystyle{ =\left{{ 2{B}^{2}- 4\left({ X- 2{A}^{2} }\right) }\right} }$

$3$\displaystyle{ 4X- 6X }$ $3$\displaystyle{ ={ 2{B}^{2}+8{A}^{2}- 2A }^{2} }$

$3$\displaystyle{ - 2X }$ $3$\displaystyle{ =6{A}^{2}+2{B}^{2} }$

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

$3$\displaystyle{ =- 3{ \left({ \begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array} }\right) }^{2}- { \left({ \begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array} }\right) }^{2} }$

$3$\displaystyle{ =- 3\left({ \begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array} }\right)\left({ \begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array} }\right) }$ $3$\displaystyle{ - \left({ \begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array} }\right)\left({ \begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =- 3\left({ \begin{array} 19 & - 10 & \vspace{6}\\ - 4 & 11 & \vspace{6}\\ \end{array} }\right)- \left({ \begin{array}7& - 27 & \vspace{6}\\ - 9 & 52 & \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =\left({ \begin{array} - 57 & 30 & \vspace{6}\\ 12 & - 33 & \vspace{6}\\ \end{array} }\right)- \left({ \begin{array}7& - 27 & \vspace{6}\\ - 9 & 52 & \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =\left({ \begin{array} - 64 & 57 & \vspace{6}\\ 21 & - 85 & \vspace{6}\\ \end{array} }\right) }$

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

(2)より

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

(3)を(1)に代入する．

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

$3$\displaystyle{ 4P }$ $3$\displaystyle{ =A+6B }$

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

(4)を(3)に代入する

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

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

(4)，(5)に $3$\displaystyle{ A }$ と $3$\displaystyle{ B }$ の具体的な行列を代入する．

$3$\displaystyle{ P }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\left({\begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right)+\frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\left({\begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array} - \frac{\hspace{2}3\hspace{2}}{\hspace{2}4\hspace{2}} &\frac{\hspace{2}5\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}&\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ \end{array}}\right)+\left({\begin{array}3& - \frac{\hspace{2}9\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} &\frac{\hspace{2}21\hspace{2}}{\hspace{2}2\hspace{2}}& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array}\frac{\hspace{2}9\hspace{2}}{\hspace{2}4\hspace{2}}& - \frac{\hspace{2}13\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ - 1 &\frac{\hspace{2}43\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ Q }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\left({\begin{array} - 3 &5& \vspace{6}\\ 2&1& \vspace{6}\\ \end{array}}\right)- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\left({\begin{array}2& - 3 & \vspace{6}\\ - 1 &7& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array} - \frac{\hspace{2}3\hspace{2}}{\hspace{2}4\hspace{2}} &\frac{\hspace{2}5\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}&\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ \end{array}}\right)- \left({\begin{array}3& - \frac{\hspace{2}9\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} &\frac{\hspace{2}21\hspace{2}}{\hspace{2}2\hspace{2}}& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array} - \frac{\hspace{2}15\hspace{2}}{\hspace{2}4\hspace{2}} &\frac{\hspace{2}23\hspace{2}}{\hspace{2}4\hspace{2}}& \vspace{6}\\ 2& - \frac{\hspace{2}41\hspace{2}}{\hspace{2}4\hspace{2}} & \vspace{6}\\ \end{array}}\right) }$

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作成：学生スタッフ

最終更新日： 2025年7月3日