行列の積の行列式に関する定理

定理

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

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

が成り立つ．

証明

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

${\displaystyle A=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)}$ ，${\displaystyle B=\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)}$

とする．

${\displaystyle b_1=\left(\begin{array}{cc}{b}_{11}& b_{12}\end{array}\right)}$，${\displaystyle b_2=\left(\begin{array}{cc}{b}_{21}& b_{12}\end{array}\right)}$

とおくと

${\displaystyle B=\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)}$

と表わせる．

${\displaystyle AB=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)}$

行列の積を参照

${\displaystyle =\left(\begin{array}{cc}{a}_{11}{b}_{11}+a_{12}{b}_{21}& a_{11}{b}_{12}+a_{12}{b}_{22}\\ a_{21}{b}_{11}+a_{22}{b}_{21}& a_{21}{b}_{12}+a_{22}{b}_{22}\end{array}\right)}$

${\displaystyle =\left(\begin{array}{c}{a}_{11}\left(\begin{array}{cc}{b}_{11}& b_{12}\end{array}\right)+a_{12}\left(\begin{array}{cc}{b}_{21}& b_{22}\end{array}\right)\\ a_{21}\left(\begin{array}{cc}{b}_{11}& b_{12}\end{array}\right)+a_{22}\left(\begin{array}{cc}{b}_{21}& b_{22}\end{array}\right)\end{array}\right)}$

${\displaystyle =\left(\begin{array}{c}{a}_{11}{b}_1+a_{12}{b}_2\\ a_{21}{b}_1+a_{22}{b}_2\end{array}\right)}$

${\displaystyle \left|AB\right|=\left|\begin{array}{c}{a}_{11}{b}_1+a_{12}{b}_2\\ a_{21}{b}_1+a_{22}{b}_2\end{array}\right|}$

行列式の和の性質を利用して行列式を分割していく．

${\displaystyle =\left|\begin{array}{c}{a}_{11}{b}_1\\ a_{21}{b}_1+a_{22}{b}_2\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_2\\ a_{21}{b}_1+a_{22}{b}_2\end{array}\right|}$

${\displaystyle =\left|\begin{array}{c}{a}_{11}{b}_1\\ a_{21}{b}_1\end{array}\right|+\left|\begin{array}{c}{a}_{11}{b}_1\\ a_{22}{b}_2\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_2\\ a_{21}{b}_1\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_2\\ a_{22}{b}_2\end{array}\right|}$

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

${\displaystyle =a_{11}{a}_{21}\left|\begin{array}{c}{b}_1\\ b_1\end{array}\right|+a_{11}{a}_{22}\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|+a_{12}{a}_{21}\left|\begin{array}{c}{b}_2\\ b_1\end{array}\right|+a_{12}{a}_{22}\left|\begin{array}{c}{b}_2\\ b_2\end{array}\right|}$

同じ行があるときの性質より， ${\displaystyle \left|\begin{array}{c}{b}_1\\ b_1\end{array}\right|=0}$，${\displaystyle \left|\begin{array}{c}{b}_2\\ b_2\end{array}\right|=0}$ となる，よって

${\displaystyle =a_{11}{a}_{22}\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|+a_{12}{a}_{21}\left|\begin{array}{c}{b}_2\\ b_1\end{array}\right|}$

sgnを使って，式を整理する．

${\displaystyle =a_{11}{a}_{22}\mathrm{sgn}\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right)\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|+a_{12}{a}_{21}\mathrm{sgn}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|}$

${\displaystyle =\left\{a_{11}{a}_{22}\mathrm{sgn}\left(\begin{array}{cc}1& 2\\ 1& 2\end{array}\right)+a_{12}{a}_{21}\mathrm{sgn}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\right\}\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|}$

${\displaystyle =\left\{{\displaystyle \sum \mathrm{sgn}\left(\begin{array}{cc}1& 2\\ i_1& i_2\end{array}\right)}{a}_{1i_1}{a}_{2i_2}\right\}\left|\begin{array}{c}{b}_1\\ b_2\end{array}\right|}$

行列式の定義より

${\displaystyle =\left|A\right|·\left|B\right|}$

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．

　

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最終更新日：2022年8月27日