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逆(方阵) 小节,第 3 段
观察这个方阵,我们如果用另一个矩阵乘$A$,则得到的结果矩阵中的每一列应该都是【$\begin{bmatrix}1\2\end{bmatrix}$的倍数】,所以我们不可能从$AB$的乘积中得到单位矩阵$I$。
应该是 $\begin{bmatrix}1\\3\end{bmatrix}$的倍数 ?
$\begin{bmatrix}1\\3\end{bmatrix}$的倍数
Gram-Schmidt正交化法 小节,倒数第 2 段
* 单位化,$q_1=\frac{1}{\sqrt 3}\begin{bmatrix}1\1\1\end{bmatrix},\quad 【q_2=\frac{1}{\sqrt 2}\begin{bmatrix}1\0\2\end{bmatrix}$】...
应该是 q_2=\frac{1}{\sqrt 2}\begin{bmatrix}0\\-1\\1\end{bmatrix}$ 。
q_2=\frac{1}{\sqrt 2}\begin{bmatrix}0\\-1\\1\end{bmatrix}$
The text was updated successfully, but these errors were encountered:
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ch03
逆(方阵) 小节,第 3 段
应该是
$\begin{bmatrix}1\\3\end{bmatrix}$的倍数
?ch17
Gram-Schmidt正交化法 小节,倒数第 2 段
应该是
q_2=\frac{1}{\sqrt 2}\begin{bmatrix}0\\-1\\1\end{bmatrix}$
。The text was updated successfully, but these errors were encountered: