A-M-n-n-A-2-A-I-A-1-

Question Number 12044 by 7991 last updated on 10/Apr/17

A \in M_{n \times n}

A^{2} = A

{(I + A)}^{- 1} = \dots . ? ? ?

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17

X = I + A

Y = I - A

Y = I^{2} - A^{2} = (I + A) (I - A) = X Y

X^{- 1} . Y = X^{- 1} . X Y = I . Y = Y

X^{- 1} . Y . Y^{- 1} = Y . Y^{- 1} = I \Rightarrow X^{- 1} = {(I + A)}^{- 1} = I