vault backup: 2026-01-17 14:36:49

pull/6/head
pjokerx 3 months ago
parent 654d1cca97
commit 4f50754479

@ -68,7 +68,7 @@ $$
**解析**
证明​
设 A为 n阶正交矩阵n≥2
$$\boldsymbol{A}^\top\boldsymbol{A}=\boldsymbol{E}$$
$$\boldsymbol{A}^T\boldsymbol{A}=\boldsymbol{E}$$
又由伴随矩阵与逆矩阵的关系:
$$\boldsymbol A^{-1} = \frac{1}{|A|}\boldsymbol{A}^*$$
联立得

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