解：由 B=(E+A)'(E-4),得 (E+A)B=E—A 从而，B十AB=E一A 即 (E+A)(E+B)=2E 故(E+B)1=与(A+E)解: 由 B = (E + A) −1 (E − A),得 (E + A)B = E − A ( ) 2 1 ) 1 E + B = A+ E 故（ −               − − − = 0 0 3 4 0 2 3 0 1 2 0 0 1 0 0 0 从而，B + AB = E − A 即 ( E + A )( E + B ) = 2E