Matrice cu elemente intregi si determinant nenul

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Cezar Lupu
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Matrice cu elemente intregi si determinant nenul

Post by Cezar Lupu »

Fie \( n\in\mathbb{N}^{*} \) si \( X, Y\in M_{n}(\mathbb{Z}) \), \( \alpha, \beta\in\mathbb{Z} \) astfel incat \( \det X,\ \det Y \) si \( \alpha +\beta \) sunt numere impare. Sa se demonstreze ca \( \det( \alpha X+\beta Y)\neq 0 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
Marius Mainea
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Post by Marius Mainea »

Presupunem \( \beta \) par si \( \alpha \) impar. Avem \( \det(\alpha X+\beta Y)=\alpha^n\det^nX+m\alpha^{n-1}\beta \det^{n-1} X\det Y+...+\beta^n\det^nY\neq0 \).
aleph
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Post by aleph »

MARIUS MAINEA wrote:Presupunem \( \beta \) par si \( \alpha \) impar. Avem \( \det(\alpha X+\beta Y)=\alpha^n\det^nX+m\alpha^{n-1}\beta \det^{n-1} X\det Y+...+\beta^n\det^nY\neq0 \).
Cam ciudată formula ...
E suficient să se lucreze modulo 2, şi gata!
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