1,a+b=ab=>ab-a-b+E=E=>(a-E)(b-E)=E ----(1)
显然|a-E|不等于0,否则|a-E||b-E|=0与(1)矛盾
2,设e-ab的逆阵为c=>(e-ab)c=e=>c-abc=e=>bc-babc=b=>ebc-babc=b
=>(e-ba)bc=b =>(e-ba)bcb^(-1)=e
3, 设a+b的逆阵为c,(a-¹+b-¹)abc=(a+b)c=e
=>a-¹+b-¹的逆阵为abc
1,a+b=ab=>ab-a-b+E=E=>(a-E)(b-E)=E ----(1)
显然|a-E|不等于0,否则|a-E||b-E|=0与(1)矛盾
2,设e-ab的逆阵为c=>(e-ab)c=e=>c-abc=e=>bc-babc=b=>ebc-babc=b
=>(e-ba)bc=b =>(e-ba)bcb^(-1)=e
3, 设a+b的逆阵为c,(a-¹+b-¹)abc=(a+b)c=e
=>a-¹+b-¹的逆阵为abc