可逆矩阵性质的证明证明:(AB)^-1 = B^-1 A^-1我的证明是(AB) (AB)^-1 = E => A[B(AB)^-1] = AA^-1=> B(AB)^-1 = A^-1=> B(AB)^-1 = EA^-1=> B(AB)^-1 = BB^-1 A^-1=> (AB)^-1 = B^-1 A^-1我的疑问是为什么不能是:(AB) (AB)^-1 = E => A[
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![可逆矩阵性质的证明证明:(AB)^-1 = B^-1 A^-1我的证明是(AB) (AB)^-1 = E => A[B(AB)^-1] = AA^-1=> B(AB)^-1 = A^-1=> B(AB)^-1 = EA^-1=> B(AB)^-1 = BB^-1 A^-1=> (AB)^-1 = B^-1 A^-1我的疑问是为什么不能是:(AB) (AB)^-1 = E => A[](/uploads/image/z/3891690-18-0.jpg?t=%E5%8F%AF%E9%80%86%E7%9F%A9%E9%98%B5%E6%80%A7%E8%B4%A8%E7%9A%84%E8%AF%81%E6%98%8E%E8%AF%81%E6%98%8E%EF%BC%9A%28AB%29%5E-1+%3D+B%5E-1+A%5E-1%E6%88%91%E7%9A%84%E8%AF%81%E6%98%8E%E6%98%AF%28AB%29+%28AB%29%5E-1+%3D+E+%3D%3E+A%5BB%28AB%29%5E-1%5D+%3D+AA%5E-1%3D%3E+B%28AB%29%5E-1+%3D+A%5E-1%3D%3E+B%28AB%29%5E-1+%3D+EA%5E-1%3D%3E+B%28AB%29%5E-1+%3D+BB%5E-1+A%5E-1%3D%3E+%28AB%29%5E-1+%3D+B%5E-1+A%5E-1%E6%88%91%E7%9A%84%E7%96%91%E9%97%AE%E6%98%AF%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E8%83%BD%E6%98%AF%EF%BC%9A%28AB%29+%28AB%29%5E-1+%3D+E+%3D%3E+A%5B)
可逆矩阵性质的证明证明:(AB)^-1 = B^-1 A^-1我的证明是(AB) (AB)^-1 = E => A[B(AB)^-1] = AA^-1=> B(AB)^-1 = A^-1=> B(AB)^-1 = EA^-1=> B(AB)^-1 = BB^-1 A^-1=> (AB)^-1 = B^-1 A^-1我的疑问是为什么不能是:(AB) (AB)^-1 = E => A[
可逆矩阵性质的证明
证明:(AB)^-1 = B^-1 A^-1
我的证明是
(AB) (AB)^-1 = E
=> A[B(AB)^-1] = AA^-1
=> B(AB)^-1 = A^-1
=> B(AB)^-1 = EA^-1
=> B(AB)^-1 = BB^-1 A^-1
=> (AB)^-1 = B^-1 A^-1
我的疑问是为什么不能是:
(AB) (AB)^-1 = E
=> A[B(AB)^-1] = AA^-1
=> B(AB)^-1 = A^-1
=> B(AB)^-1 = A^-1E
=> B(AB)^-1 = A^-1 B^-1 B
=> (AB)^-1 = A^-1 B^-1
尽管我知道(AB)^-1 = A^-1 B^-1 是错的
可逆矩阵性质的证明证明:(AB)^-1 = B^-1 A^-1我的证明是(AB) (AB)^-1 = E => A[B(AB)^-1] = AA^-1=> B(AB)^-1 = A^-1=> B(AB)^-1 = EA^-1=> B(AB)^-1 = BB^-1 A^-1=> (AB)^-1 = B^-1 A^-1我的疑问是为什么不能是:(AB) (AB)^-1 = E => A[
(AB) (AB)^-1 = E
=> A[B(AB)^-1] = AA^-1
=> B(AB)^-1 = A^-1
=> B(AB)^-1 = A^-1E
=> B(AB)^-1 = A^-1 B^-1 B
以上正确,以下不正确,因为矩阵不满足交换律,上面等式中的B不能约去.
=> (AB)^-1 = A^-1 B^-1
B(AB)^-1 = A^-1 B^-1 B
=> (AB)^-1 = A^-1 B^-1
这一步是不成立的,你的依据是什么?
B(AB)^-1 = BB^-1 A^-1
=> (AB)^-1 = B^-1 A^-1
上面这个是两边同乘以 B^-1得到的