矩陣,A=2I+B和AB=O,若(A+I)^8=kA+I,求k=?
已知 A=2I+B 和 AB=O, 若(A+I)^8=kA+I 求k 感恩 \(A=2I+B \Rightarrow B = A-2I\)\(\Rightarrow AB = A^2-2A=0\)
\(\Rightarrow A^2 = 2A \Rightarrow A^n = 2^{n-1} A,\,\forall n\in\mathbb{N}\)
\(\displaystyle\left(A+I\right)^8 = C^8_0 A^8 + C^8_1 A^7+\cdots+C^8_7A + C^8_8 I\)
\(\displaystyle= 2^7 C^8_0A + 2^6 C^8_1 A+\cdots + C^8_7 A + I\)
\(\displaystyle= \frac{1}{2}\left(2^8 C^8_0 + 2^7 C^8_1+\cdots 2C^8_7 +C^8_8 - 1\right) A +I\)
\(\displaystyle= \frac{1}{2}\left(3^8 - 1\right) A +I\)
\(\displaystyle= 3280 A+I\) [quote]原帖由 [i]weiye[/i] 於 2016-7-3 08:50 PM 發表 [url=https://math.pro/db/redirect.php?goto=findpost&pid=15870&ptid=2551][img]https://math.pro/db/images/common/back.gif[/img][/url]
\(A=2I+B \Rightarrow B = A-2I\)
\(\Rightarrow AB = A^2-2A=0\)
\(\Rightarrow A^2 = 2A \Rightarrow A^n = 2^{n-1} A,\,\forall n\in\mathbb{N}\)
... [/quote]
感謝老師.....
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