证明如下:利用逆矩阵的定义来证明!
![](https://iknow-pic.cdn.bcebos.com/35a85edf8db1cb130ad64573cd54564e92584b16?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
因为,A,B,A+B均可逆,
那么可设它的逆为C矩阵,
同时E为单位矩阵,则有
(A+B)C=E, C(A+B)=E
而 (A+B)B⁻¹[A⁻¹+B⁻¹]⁻¹A⁻¹
=[AB⁻¹+E]{A[A⁻¹+B⁻¹]}⁻¹
=[E+AB⁻¹][E+AB⁻¹]⁻¹ =E
同时, B⁻¹[A⁻¹+B⁻¹]⁻¹A⁻¹(A+B)
={[A⁻¹+B⁻¹]B}⁻¹[E+A⁻¹B]
=[A⁻¹B+E]⁻¹[A⁻¹B+E] =E
所以(A+B)⁻¹=B⁻¹[A⁻¹+B⁻¹]⁻¹A⁻¹
得证!