# 行列の積の行列式に関する定理

## 定理

$A$$B$$n$次の正方行列とするとき，

$\left|AB\right|=\left|A\right|·\left|B\right|$

が成り立つ．

## 証明

### ●$A$ ，$B$が$2$次の正方行列の場合

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

とする．

${b}_{1}=\left(\begin{array}{cc}{b}_{11}& {b}_{12}\end{array}\right)$${b}_{2}=\left(\begin{array}{cc}{b}_{21}& {b}_{12}\end{array}\right)$

とおくと

$B=\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right)$

と表わせる．

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

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

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

$=\left(\begin{array}{c}{a}_{11}{b}_{1}+{a}_{12}{b}_{2}\\ {a}_{21}{b}_{1}+{a}_{22}{b}_{2}\end{array}\right)$

$\left|AB\right|=\left|\begin{array}{c}{a}_{11}{b}_{1}+{a}_{12}{b}_{2}\\ {a}_{21}{b}_{1}+{a}_{22}{b}_{2}\end{array}\right|$

$=\left|\begin{array}{c}{a}_{11}{b}_{1}\\ {a}_{21}{b}_{1}+{a}_{22}{b}_{2}\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_{2}\\ {a}_{21}{b}_{1}+{a}_{22}{b}_{2}\end{array}\right|$

$=\left|\begin{array}{c}{a}_{11}{b}_{1}\\ {a}_{21}{b}_{1}\end{array}\right|+\left|\begin{array}{c}{a}_{11}{b}_{1}\\ {a}_{22}{b}_{2}\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_{2}\\ {a}_{21}{b}_{1}\end{array}\right|+\left|\begin{array}{c}{a}_{12}{b}_{2}\\ {a}_{22}{b}_{2}\end{array}\right|$

$={a}_{11}{a}_{21}\left|\begin{array}{c}{b}_{1}\\ {b}_{1}\end{array}\right|+{a}_{11}{a}_{22}\left|\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right|+{a}_{12}{a}_{21}\left|\begin{array}{c}{b}_{2}\\ {b}_{1}\end{array}\right|+{a}_{12}{a}_{22}\left|\begin{array}{c}{b}_{2}\\ {b}_{2}\end{array}\right|$

$={a}_{11}{a}_{22}\left|\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right|+{a}_{12}{a}_{21}\left|\begin{array}{c}{b}_{2}\\ {b}_{1}\end{array}\right|$

sgnを使って，式を整理する．

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

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

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

$=\left|A\right|·\left|B\right|$

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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