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

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

How to find B^(-1)B^(-1)?;We know that B^2=B+2I_3B^2=B+2I_3