How to find #B^(-1)#?;We know that #B^2=B+2I_3#

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Answer 1 · 2017-04-02T19:41:54

#B^{-1}=(1/2)B-(1/2)I_3#.

You have

#B^2=B+2I_3#.

Multiply by #B^{-1}#:

#B=I_3+2B#^{-1}#

And then from simple algebra:

#(1/2)B-(1/2)I_3=B^{-1}#.

Answer 2 · 2017-04-02T22:55:58

#B^-1=1/2(B-I_3)#

Any matrix #B# obeys it's characteristic polynomial

Given #B=((0,1,1),(1,0,1),(1,1,0))# we have

#B^3-3B-2I_3=0_3# because

#p(lambda)=lambda^3-3lambda-2#

Now multiplying by left this relationship we have

#B^-1B^3-3B^-1B-2B^-1=0_3# or

#B^2-3I_3-2B^-1=0_3# then

#2B^-1=B^2-3I_3# but

#B^2=B+2I_3# so finally

#B^-1=1/2(B-I_3)#