Question

Prove by induction that \[(A_1A_2...A_n)^{-1}={A_n}^{-1}...{A_2}^{-1}{A_1}^{-1} \]\[\forall n\in \mathbb{N}\] and A_i for i=1, 2, ... are invertible matrices of same dimensions

Answer 1

Base step: Let n=1 \[(A_1)^{-1}=A_1^{-1}\]\[A_1^{-1}=A_1^{-1}\]
Inductive Hyp. : Assume the statement holds true for n=m for some m in Natural Numbers.
Inductive Step: Let n=m+1\[(A_1A_2...A_mA_{m+1})^{-1}=A_{m+1}^{-1}A_m^{-1}...A_2^{-1}A_1^{-1}\]\[(A_1A_2...A_mA_{m+1})^{-1}=A_{m+1}^{-1}(A_1A_2...A_m)^{-1}\]

That's about as far as I got. Something doesn't seem right, and the Inductive Step isn't making sense as I don't know how to continue.

Answer 2

*

Answer 3

write