行列の計算

次の行列の計算をせよ．
1. $3$\displaystyle{ $ \begin{array}1&2& \vspace{6}\\ 3&4& \vspace{6}\\ \end{array} $ + $ \begin{array}5&6& \vspace{6}\\ 7&8& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
2. $3$\displaystyle{ $ \begin{array} \begin{array}3&7&8& \vspace{6}\\ \end{array} & \vspace{6}\\ \begin{array}1&5&2& \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ + $ \begin{array} \begin{array}6&2&4& \vspace{6}\\ \end{array} & \vspace{6}\\ \begin{array}7&1&9& \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ ⇒ 解答
3. $3$\displaystyle{ $ \begin{array}a&b& \vspace{6}\\ c&d& \vspace{6}\\ \end{array} $ $ \begin{array}x& \vspace{6}\\ y& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
4. $3$\displaystyle{ $ \begin{array}1&3& \vspace{6}\\ 5&7& \vspace{6}\\ \end{array} $ $ \begin{array}8&6& \vspace{6}\\ 4&2& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
5. $3$\displaystyle{ $ \begin{array}2&3& \vspace{6}\\ 4&5& \vspace{6}\\ \end{array} $ $ \begin{array} \begin{array}6&5&4& \vspace{6}\\ \end{array} & \vspace{6}\\ \begin{array}3&2&1& \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ ⇒ 解答
6. $3$\displaystyle{ $ \begin{array}a&b&c& \vspace{6}\\ \end{array} $ $ \begin{array}x& \vspace{6}\\ y& \vspace{6}\\ z& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
7. $3$\displaystyle{ $ \begin{array}a& \vspace{6}\\ b& \vspace{6}\\ c& \vspace{6}\\ \end{array} $ $ \begin{array}x&y&z& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
8. $3$\displaystyle{ $ \begin{array}2&6& \vspace{6}\\ 4&3& \vspace{6}\\ \end{array} $ - $ \begin{array}3&4& \vspace{6}\\ 1&5& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
9. $3$\displaystyle{3 $ \begin{array} \begin{array}4& \vspace{6}\\ 0& \vspace{6}\\ \end{array} & \begin{array}2& \vspace{6}\\ 5& \vspace{6}\\ \end{array} & \begin{array} - 3 & \vspace{6}\\ - 1 & \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ $3$\displaystyle{- 2 $ \begin{array} \begin{array}3& \vspace{6}\\ 4& \vspace{6}\\ \end{array} & \begin{array}1& \vspace{6}\\ - 1 & \vspace{6}\\ \end{array} & \begin{array} - 2 & \vspace{6}\\ 0& \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ ⇒ 解答
10. $3$\displaystyle{ 3 $ \begin{array} \begin{array} - 1 &2&1& \vspace{6}\\ 1&3& - 1 & \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ +2 $ \begin{array} \begin{array} - 1 &1&2& \vspace{6}\\ - 2 &1& - 5 & \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ ⇒ 解答
11. $3$\displaystyle{ $ \begin{array}3&1& - 2 & \vspace{6}\\ \end{array} $ $ \begin{array}2& \vspace{6}\\ - 3 & \vspace{6}\\ 4& \vspace{6}\\ \end{array} $ }$ ⇒ 解答
12. $3$\displaystyle{ $ \begin{array}2&3& - 2 & \vspace{6}\\ 0& - 2 &5& \vspace{6}\\ 1&1&3& \vspace{6}\\ \end{array} $ $ \begin{array} \begin{array}2& \vspace{6}\\ - 1 & \vspace{6}\\ 2& \vspace{6}\\ \end{array} & \begin{array}1& \vspace{6}\\ 5& \vspace{6}\\ - 2 & \vspace{6}\\ \end{array} & \vspace{6}\\ \end{array} $ }$ ⇒ 解答
13. $3$\displaystyle{{ $ \begin{array}1&0& - 1 & \vspace{6}\\ 1&0& - 2 & \vspace{6}\\ 3& - 2 &1& \vspace{6}\\ \end{array} $ }^{2}}$ ⇒ 解答
掃き出し法を用いて，次の連立方程式を解け．
- (*) $3$\displaystyle{ \{ \begin{array} x+3y+2z=0 & \vspace{6}\\ x- 4y+2z=7 & \vspace{6}\\ 2x- y+5z=6 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{\left{{ \begin{array} x- 2y+4z=8 & \vspace{6}\\ 3x- y+2z=9 & \vspace{6}\\ 2x+y- z=6 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{ \{ \begin{array} x- 3y+z=- 2 & \vspace{6}\\ - 3x+4y+2z=- 19 & \vspace{6}\\ 2x+y- z=11 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{ \left{{\begin{array} a- b- 2c+2d=4 & \vspace{6}\\ 2a+b- 3c- d=- 3 & \vspace{6}\\ - a+2b+2c- 2c=- 6 & \vspace{6}\\ 3a- 3b+c+d=7 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{ \left{{\begin{array} a+b- 2c- 2d=- 5 & \vspace{6}\\ 2a- 2b+c- d=8 & \vspace{6}\\ - a+b- c+2d=- 5 & \vspace{6}\\ - 2a- b+c- 3d=2 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{\displaystyle{\left{{\begin{array} a+ 2b+ d- e=- 4 & \vspace{6}\\ - a- b- c+ d=3 & \vspace{6}\\ - 2b+ 2c+ d- e=- 1 & \vspace{6}\\ a+ b+ c- 2d+ e=- 2 & \vspace{6}\\ a- b- 2c+ 2e=9 & \vspace{6}\\ \end{array}}}$ ⇒ 解答
- (*) $3$\displaystyle{ \{ \begin{array} x+3y+5z=6 & \vspace{6}\\ 3x+7y+11z=16 & \vspace{6}\\ - x+4y+7z=- 3 & \vspace{6}\\ \end{array} }$ ⇒ 解答
- (*) $3$\displaystyle{ \{ \begin{array} 3x- y- 2z=2 & \vspace{6}\\ 2y- z=- 1 & \vspace{6}\\ 3x- 5y=3 & \vspace{6}\\ \end{array} }$ ⇒ 解答
次の等式を満たす行列を求めよ．
1. $3$\displaystyle{ A=\left({\begin{array} - 2 &1& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array}}\right) }$ ， $3$\displaystyle{ B=\left({\begin{array}1&2& \vspace{6}\\ - 6 &9& \vspace{6}\\ \end{array}}\right) }$ のとき，次の行列を求めよ．
  [i] 等式 $3$\displaystyle{ A- 2X=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\left{{ 4A- 3\left({ B+3X }\right) }\right} }$ を満たす行列 $3$\displaystyle{X}$
  
  [ii] 2つの等式 $3$\displaystyle{ P+Q=A }$ ， $3$\displaystyle{ P- Q=B }$ を同時に満たす2次の正方行列 $3$\displaystyle{P}$ ， $3$\displaystyle{Q}$ ⇒ 解答
2. $3$\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 }$ $3$\displaystyle{ =\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 }$ ， $3$\displaystyle{P=A+B}$ を同時に満たす2次の正方行列 $3$\displaystyle{P}$ ， $3$\displaystyle{Q}$ ⇒ 解答
3. $3$\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
4. $3$\displaystyle{A=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} $ \begin{array}9&1& \vspace{6}\\ - 6 & - 1 & \vspace{6}\\ \end{array} $ }$ ， $3$\displaystyle{B= $ \begin{array}3&1& \vspace{6}\\ - 5 & - 3 & \vspace{6}\\ \end{array} $ }$ のとき， $3$\displaystyle{A{B}^{ 5 }{A}^{ - 1 }}$ を求めよ． ⇒ 解答
連立方程式
　　 $3$\displaystyle{ \{ \begin{array} 2x+y=7 & \vspace{6}\\ 5x+2y=16 & \vspace{6}\\ \end{array} }$
を係数行列の逆行列を利用して解け．計算過程を書くこと． ⇒解答
掃き出し法を用いて
　　 $3$\displaystyle{A= $ \begin{array}0&0&1& \vspace{6}\\ 0&1&2& \vspace{6}\\ 1&3&4& \vspace{6}\\ \end{array} $ }$
の逆行列 $3$\displaystyle{{A}^{ - 1 }}$ を求めよ． ⇒解答
平面 $3$\displaystyle{2x- 3y+z=1}$ と平面 $3$\displaystyle{3x+2y- 3z=1}$ の交線の方程式を求めよ． ⇒解答

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最終更新日：2025年11月9日