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{{Лемма|about = *|statement=<tex> \mathcal{A}^{\wedge_p} {e_{i_1}} \land {e_{i_2}} \land ... \land {e_{i_p}} \stackrel{\mathrm{def}}{=} \mathcal{A}{e_{i_1}} \land \mathcal{A}{e_{i_2}} \land... \land \mathcal{A}{e_{i_p}} </tex>}} {{Лемма|about = **|statement=Если <tex> {x_1} \land {x_2} \land... \land {x_p} \in {\wedge_p} </tex>, то <tex> \mathcal{A}^{\wedge_p} {x_1} \land {x_2} \land ... \land {x_p} = \mathcal{A}{x_1} \land \mathcal{A}{x_2} \land... \land \mathcal{A}{x_p} </tex>}} {{Лемма|about = ***|statement=<tex> \mathcal{A}^{\wedge_n} z = \det \mathcal{A} \cdot z </tex>}} ==Теорема умножения определителей =={{Теорема|statement=Пусть <tex>\mathcal{A}</tex>, <tex>\mathcal{B} \colon X \to X</tex> (автоморфизм). <br> Тогда <tex>\det (\mathcal{A} \cdot \mathcal{B}) = \det \mathcal{A} \cdot \det \mathcal{B}</tex>|proof =<tex>\det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = </tex><br><tex>(\mathcal{A} \cdot \mathcal{B})^{\wedge_n}{e_1} \land {e_2} \land... \land{e_n} \stackrel{\mathrm{(*)}}{=}</tex><br><tex>(\mathcal{A} \cdot \mathcal{B}) {e_1} \land (\mathcal{A} \cdot \mathcal{B}) {e_2} \land ... \land (\mathcal{A} \cdot \mathcal{B}) {e_n} \stackrel{\mathrm{(def\mathcal{A} \cdot \mathcal{B})}}{=}</tex><br><tex>\mathcal{A} (\mathcal{B} {e_1}) \land \mathcal{A} (\mathcal{B} {e_2}) \land ... \land \mathcal{A} (\mathcal{B} {e_n}) \stackrel{\mathrm{(**)}}{=}</tex><br><tex>\mathcal{A}^{\wedge_n}(\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n})\stackrel{\mathrm{(***)}}{=}</tex><br><tex>\det \mathcal{A} \cdot (\mathcal{B} {e_1} \land \mathcal{B} {e_2} \land ... \land \mathcal{B} {e_n}) \stackrel{\mathrm{(***)}}{=}</tex><br><tex>\det \mathcal{A} \cdot \mathcal{B}^{\wedge_n}({e_1} \land {e_2} \land ... \land {e_n}) = </tex><br><tex>\det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n} </tex><br> т.е. <tex> \det (\mathcal{A} \cdot \mathcal{B}) {e_1} \land {e_2} \land... \land{e_n} = </tex><br><tex>\det \mathcal{A} \cdot \det \mathcal{B} \cdot {e_1} \land {e_2} \land ... \land {e_n}</tex> }}{{Утверждение|statement='''Следствие 1: '''<br>Пусть <tex>\mathcal{A} \colon X \to X</tex> обратим, т.е. <tex>(\exists \mathcal{A}^{-1}) </tex><br><tex>\det \mathcal{A} \ne 0 </tex><br>Тогда: <tex>\mathcal{A}^{-1} = (\det \mathcal{A})^{-1} </tex>|proof=<tex>\mathcal{A}^{-1} \cdot \mathcal{A} = \mathcal{I} \det (\mathcal{A}^{-1} \cdot \mathcal{A}) = </tex><br><tex>\det \mathcal{A}^{-1} \cdot \det \mathcal{A} = \det \mathcal{I} = 1 (\det {E} = 1) </tex><br>Итого: <tex> \det \mathcal{A}^{-1} = (\det \mathcal{A})^{-1} </tex> }}{{Утверждение|statement='''Следствие 2: '''<br>Пусть <tex>\tilde{A} = T^{-1}AT</tex> <br>Тогда <tex> \det \tilde{A} = \det A </tex> <br><tex> \det \tilde{A} = \det T \det A \det T^{-1} = \det A </tex> }}
