行列の積の行列式に関する定理

定理

$3$\displaystyle{A}$ ， $3$\displaystyle{B}$ を $3$\displaystyle{n}$ 次の正方行列とするとき，

$3$\displaystyle{ \left|{ AB }\right|=\left|{A}\right|\cdot \left|{B}\right| }$

が成り立つ．

証明

● $3$\displaystyle{A}$ ， $3$\displaystyle{B}$ が $3$\displaystyle{2}$ 次の正方行列の場合

$3$\displaystyle{A=\left({ \begin{array} \hspace{1}{a}_{ 11 } & \hspace{1}{a}_{ 12 } & \vspace{6}\\ \hspace{1}{a}_{ 21 } & \hspace{1}{a}_{ 22 } & \vspace{6}\\ \end{array} }\right)}$ ， $3$\displaystyle{B=\left({ \begin{array} \hspace{1}{b}_{ 11 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \hspace{1}{b}_{ 21 } & \hspace{1}{b}_{ 22 } & \vspace{6}\\ \end{array} }\right)}$

とする．

$3$\displaystyle{\hspace{1}{\displaystyle{b}}_{1}=\left({ \begin{array} \hspace{1}{b}_{ 11 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \end{array} }\right)}$ ， $3$\displaystyle{\hspace{1}{\displaystyle{b}}_{2}=\left({ \begin{array} \hspace{1}{b}_{ 21 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \end{array} }\right)}$

とおくと

$3$\displaystyle{B=\left({ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right)}$

と表わせる．

$3$\displaystyle{AB=\left({ \begin{array} \hspace{1}{a}_{ 11 } & \hspace{1}{a}_{ 12 } & \vspace{6}\\ \hspace{1}{a}_{ 21 } & \hspace{1}{a}_{ 22 } & \vspace{6}\\ \end{array} }\right)\left({ \begin{array} \hspace{1}{b}_{ 11 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \hspace{1}{b}_{ 21 } & \hspace{1}{b}_{ 22 } & \vspace{6}\\ \end{array} }\right)}$

行列の積を参照

$3$\displaystyle{=\left({ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{b}_{ 11 }+\hspace{1}{a}_{ 12 }\hspace{1}{b}_{ 21 } & \hspace{1}{a}_{ 11 }\hspace{1}{b}_{ 12 }+\hspace{1}{a}_{ 12 }\hspace{1}{b}_{ 22 } & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{b}_{ 11 }+\hspace{1}{a}_{ 22 }\hspace{1}{b}_{ 21 } & \hspace{1}{a}_{ 21 }\hspace{1}{b}_{ 12 }+\hspace{1}{a}_{ 22 }\hspace{1}{b}_{ 22 } & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\left({ \begin{array} \hspace{1}{a}_{ 11 }\left({ \begin{array} \hspace{1}{b}_{ 11 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \end{array} }\right)+\hspace{1}{a}_{ 12 }\left({ \begin{array} \hspace{1}{b}_{ 21 } & \hspace{1}{b}_{ 22 } & \vspace{6}\\ \end{array} }\right) & \vspace{6}\\ \hspace{1}{a}_{ 21 }\left({ \begin{array} \hspace{1}{b}_{ 11 } & \hspace{1}{b}_{ 12 } & \vspace{6}\\ \end{array} }\right)+\hspace{1}{a}_{ 22 }\left({ \begin{array} \hspace{1}{b}_{ 21 } & \hspace{1}{b}_{ 22 } & \vspace{6}\\ \end{array} }\right) & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{=\left({ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 12 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right)}$

$3$\displaystyle{\left|{ AB }\right|=\left|{ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 12 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

行列式の和の性質を利用して行列式を分割していく．

$3$\displaystyle{=\left|{ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|+\left|{ \begin{array} \hspace{1}{a}_{ 12 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1}+\hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

$3$\displaystyle{=\left|{ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|+\left|{ \begin{array} \hspace{1}{a}_{ 11 }\hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|+\left|{ \begin{array} \hspace{1}{a}_{ 12 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{a}_{ 21 }\hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|+\left|{ \begin{array} \hspace{1}{a}_{ 12 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{a}_{ 22 }\hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

定数倍の性質を利用して，行列 $3$\displaystyle{A}$ の成分をくくりだす．

$3$\displaystyle{=\hspace{1}{a}_{ 11 }\hspace{1}{a}_{ 21 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|+\hspace{1}{a}_{ 11 }\hspace{1}{a}_{ 22 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|+\hspace{1}{a}_{ 12 }\hspace{1}{a}_{ 21 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|+\hspace{1}{a}_{ 12 }\hspace{1}{a}_{ 22 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

同じ行があるときの性質より， $3$\displaystyle{\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|=0}$ ， $3$\displaystyle{\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|=0}$ となる，よって

$3$\displaystyle{=\hspace{1}{a}_{ 11 }\hspace{1}{a}_{ 22 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|+\hspace{1}{a}_{ 12 }\hspace{1}{a}_{ 21 }\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \end{array} }\right|}$

sgnを使って，式を整理する．

$3$\displaystyle{=\hspace{1}{a}_{ 11 }\hspace{1}{a}_{ 22 }sgn\left({ \begin{array}1&2& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array} }\right)\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|+\hspace{1}{a}_{ 12 }\hspace{1}{a}_{ 21 }sgn\left({ \begin{array}2&1& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array} }\right)\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

$3$\displaystyle{=\left{{ \hspace{1}{a}_{ 11 }\hspace{1}{a}_{ 22 }sgn\left({ \begin{array}1&2& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array} }\right)+\hspace{1}{a}_{ 12 }\hspace{1}{a}_{ 21 }sgn\left({ \begin{array}2&1& \vspace{6}\\ 1&2& \vspace{6}\\ \end{array} }\right) }\right}\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

$3$\displaystyle{=\left{{ \displaystyle{\sum sgn\left({ \begin{array}1&2& \vspace{6}\\ \hspace{1}{i}_{1} & \hspace{1}{i}_{2} & \vspace{6}\\ \end{array} }\right) }\hspace{1}{a}_{ 1\hspace{1}{i}_{1} }\hspace{1}{a}_{ 2\hspace{1}{i}_{2} } }\right}\left|{ \begin{array} \hspace{1}{\displaystyle{b}}_{1} & \vspace{6}\\ \hspace{1}{\displaystyle{b}}_{2} & \vspace{6}\\ \end{array} }\right|}$

行列式の定義より

$3$\displaystyle{=\left|{A}\right|\cdot \left|{B}\right|}$

となり，証明された．

2次より次数が大きい場合も同様にして証明できる．

ホーム>>カテゴリー分類>>行列>>線形代数>>行列の積の行列式に関する定理

最終更新日：2022年8月27日

行列の積の行列式に関する定理

定理

証明

● ，が次の正方行列の場合

● $3$\displaystyle{A}$ ， $3$\displaystyle{B}$ が $3$\displaystyle{2}$ 次の正方行列の場合