A³=2E
B=A³-2A²+2A=2E+2A-2A²
AB=2A+2A²-2A³)=-4E+2A+2A²
A²B=2A(-2E+A+A²)=(-4A+2A²+2A³)=4E-4A+2A²
(E+αA+βA²)B=(2-4α+4β)E+(2+2α-4β)A+(-2+2α+2β)A²
令2+2α-4β=-2+2α+2β=0 得α=0.5 ,β=0.5
所以(E+0.5A+0.5A²)B=2E
B可逆,且B^(-1)=0.5E+0.25A+0.25A²=(2E+A+A²)/4
A³=2E
B=A³-2A²+2A=2E+2A-2A²
AB=2A+2A²-2A³)=-4E+2A+2A²
A²B=2A(-2E+A+A²)=(-4A+2A²+2A³)=4E-4A+2A²
(E+αA+βA²)B=(2-4α+4β)E+(2+2α-4β)A+(-2+2α+2β)A²
令2+2α-4β=-2+2α+2β=0 得α=0.5 ,β=0.5
所以(E+0.5A+0.5A²)B=2E
B可逆,且B^(-1)=0.5E+0.25A+0.25A²=(2E+A+A²)/4