基本的な行列の問題

■問題

$3$\displaystyle{ A=\left({\begin{array}5&3& \vspace{6}\\ - 7 & - 10 & \vspace{6}\\ \end{array}}\right) }$ ， $3$\displaystyle{ B=\left({\begin{array} - 2 &5& \vspace{6}\\ 1& - 9 & \vspace{6}\\ \end{array}}\right) }$ のとき，次の行列を求めよ．

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

[ii]2つの等式 $3$\displaystyle{ PQ=A }$ …(1) ， $3$\displaystyle{ P=A+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．

■解答

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

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

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

$3$\displaystyle{ 12A- 20X }$ $3$\displaystyle{ =- 10A- 10X+15B }$

$3$\displaystyle{ - 10X }$ $3$\displaystyle{ =- 22A+15B }$

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

$3$\displaystyle{ =\frac{\hspace{2}11\hspace{2}}{\hspace{2}5\hspace{2}}\left({\begin{array}5&3& \vspace{6}\\ - 7 & - 10 & \vspace{6}\\ \end{array}}\right)- \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}}\left({\begin{array} - 2 &5& \vspace{6}\\ 1& - 9 & \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array}11&\frac{\hspace{2}33\hspace{2}}{\hspace{2}5\hspace{2}}& \vspace{6}\\ - \frac{\hspace{2}77\hspace{2}}{\hspace{2}5\hspace{2}} & - 22 & \vspace{6}\\ \end{array}}\right)+\left({\begin{array}3& - \frac{\hspace{2}15\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} &\frac{\hspace{2}27\hspace{2}}{\hspace{2}2\hspace{2}}& \vspace{6}\\ \end{array}}\right) }$

$3$\displaystyle{ =\left({\begin{array}14& - \frac{\hspace{2}9\hspace{2}}{\hspace{2}10\hspace{2}} & \vspace{6}\\ - \frac{\hspace{2}169\hspace{2}}{\hspace{2}10\hspace{2}} & - \frac{\hspace{2}17\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \end{array}}\right) }$

[ii]2つの等式 $3$\displaystyle{ PQ=A }$ …(1) ， $3$\displaystyle{ P=A+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．

(1)より

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

(2)より

$3$\displaystyle{ Q }$ $3$\displaystyle{ ={P}^{ - 1 }A }$

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

逆行列の計算をする．

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} - 9 \hspace{2}}\left({ \begin{array} - 19 & - 8 & \vspace{6}\\ 6&3& \vspace{6}\\ \end{array} }\right)\left({ \begin{array}5&3& \vspace{6}\\ - 7 & - 10 & \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}\left({ \begin{array} - 19 & - 8 & \vspace{6}\\ 6&3& \vspace{6}\\ \end{array} }\right)\left({ \begin{array}5&3& \vspace{6}\\ - 7 & - 10 & \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}9\hspace{2}}\left({ \begin{array} - 39 & 23 & \vspace{6}\\ 9& - 12 & \vspace{6}\\ \end{array} }\right) }$

$3$\displaystyle{ =\left({ \begin{array} \frac{\hspace{2} 13 \hspace{2}}{\hspace{2}3\hspace{2}} & - \frac{\hspace{2} 23 \hspace{2}}{\hspace{2}9\hspace{2}} & \vspace{6}\\ - 1 & - \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}} & \vspace{6}\\ \end{array} }\right) }$

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作成：学生スタッフ

最終更新日： 2022年6月16日