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要用 二項式定理 前,要先確認 乘法交換律 是否存在~

(思考: \(\left(x+y\right)^2=\left(x+y\right)\cdot\left(x+y\right)= x \cdot x+ \underline{x \cdot y} + \underline{y \cdot x} + y \cdot y = x \cdot x+ \underline{x \cdot y} + \underline{x \cdot y} + y \cdot y =x^2+\underline{2xy}+y^2 \) )


\(\left[\begin{array}{cc}x & 0 \\ 0 & y\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & x \\ 0 & 0\end{array}\right]\)

\(\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{cc}x & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{cc}0 & y \\ 0 & 0\end{array}\right]\)

\(\Rightarrow \left[\begin{array}{cc}x & 0 \\ 0 & y\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]\neq\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]\left[\begin{array}{cc}x & 0 \\ 0 & y\end{array}\right]\)



至於正確解答,可以經由觀察出規律,然後透過數學歸納法證明。

(\(\left[\begin{array}{cc}x & 1 \\ 0 & y\end{array}\right]^2=\left[\begin{array}{cc}x^2 & x+y \\ 0 & y^2\end{array}\right]\)

 \(\left[\begin{array}{cc}x & 1 \\ 0 & y\end{array}\right]^3=\left[\begin{array}{cc}x^3 & x^2+xy+y^2 \\ 0 & y^3\end{array}\right]\)

 \(\left[\begin{array}{cc}x & 1 \\ 0 & y\end{array}\right]^4=\left[\begin{array}{cc}x^4 & x^3+x^2y+xy^2+y^3 \\ 0 & y^4\end{array}\right]\))

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