Let #p# be a non singular matrix #1+p+p^2+p^3+cdots+p^n=O# (#O# denotes the null matrix), then #p^-1# is?

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Answer 1 · 2017-10-25T14:55:29

The answer is #=-(I+p+.........p^(n-1))#

We know that

#p^-1p=I#

#I+p+p^2+p^3.....p^n=O#

Multiply both sides by #p^-1#

#p^-1*(1+p+p^2+p^3.....p^n)=p^-1*O#

#p^-1*1+p^-1*p+p^-1*p^2+......p^-1*p^n=O#

#p^-1+(p^-1p)+(p^-1*p*p)+.........(p^-1p*p^(n-1))=O#

#p^-1+(I)+(I*p)+.........(I*p^(n-1))=O#

Therefore,

#p^-1=-(I+p+.........p^(n-1))#

Answer 2 · 2017-10-25T14:58:16

See below.

#p(p^-1+p+p^2+ cdots + p^(n-1))=0# but #p# by hypothesis is non singular then exists #p^-1# so

#p^-1 p(p^-1+p+p^2+ cdots + p^(n-1))=p^-1+p+p^2+ cdots + p^(n-1)=0#

and finally

#p^- 1 = - sum_(k=1)^ (n-1) p^k#

Also can be solved as

#p^-1 = -p(sum_(k=0)^(n-2)p^k) = p(p^(n-1)+p^n) = p^n(1-p)#