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矩阵的秩

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溦澜居士 Posted 2024-4-22 00:07 From mobile phone |Read mode
Last edited by hbghlyj 2025-4-27 02:47设矩阵 $A=\left(\begin{array}{rccccc}
1^{2024} & 2^{2024} & 3^{2024} & \ldots & 4047^{2024} & 4048^{2024} \\
2^{2024} & 3^{2024} & 4^{2024} & \ldots & 4048^{2024} & 4049^{2024} \\
3^{2024} & 4^{2024} & 5^{2024} & \ldots & 4049^{2024} & 4050^{2024} \\
\ldots 0^{2024} & 4048^{2024} & 4049^{2024} & \ldots & 8093^{2024} & 8094^{2024} \\
4047^{2024} & 4049^{2024} & 4050^{2024} & \ldots & 8094^{2024} & 8095^{2024}
\end{array}\right)$,则 $\operatorname{rank}(A)=$

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Czhang271828 Posted 2024-4-24 20:06
提示: 用二项式展开计算. 例如
$$
\small \begin{pmatrix}
1^3&2^3&3^3&4^3&5^3&6^3\\
2^3&3^3&4^3&5^3&6^3&7^3\\
3^3&4^3&5^3&6^3&7^3&8^3\\
4^3&5^3&6^3&7^3&8^3&9^3\\
5^3&6^3&7^3&8^3&9^3&10^3\\
6^3&7^3&8^3&9^3&10^3&11^3\\
\end{pmatrix}=
\begin{pmatrix}
1^3&3\cdot 1^2&3\cdot 1^1&1^0\\
2^3&3\cdot 2^2&3\cdot 2^1&2^0\\
3^3&3\cdot 3^2&3\cdot 3^1&3^0\\
4^3&3\cdot 4^2&3\cdot 4^1&4^0\\
5^3&3\cdot 5^2&3\cdot 5^1&5^0\\
6^3&3\cdot 6^2&3\cdot 6^1&6^0\\
\end{pmatrix}\cdot \begin{pmatrix}
0^0&1^0&2^0&3^0&4^0&5^0\\
0^1&1^1&2^1&3^1&4^1&5^1\\
0^2&1^2&2^2&3^2&4^2&5^2\\
0^3&1^3&2^3&3^3&4^3&5^3\\
\end{pmatrix}.
$$

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2025-6-5 01:49 GMT+8

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