## PhoenixFire Group Title 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 2 years ago 2 years ago

1. PhoenixFire

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.

2. UnkleRhaukus

*

3. experimentX

write |dw:1349805212895:dw|

4. experimentX

|dw:1349805366358:dw|

5. asnaseer

I would first show that: $$(A_1A_2)^{-1}=A_2^{-1}A_1^{-1}$$ Then I would assume it is true for n=k., i.e.: $$(A_1A_2...A_k)^{-1}=A_k^{-1}..A_2^{-1}A_1^{-1}$$ And finally I would show that this also leads to: $$(A_1A_2...A_k A_{k+1})^{-1}=A_{k+1}^{-1}A_k^{-1}..A_2^{-1}A_1^{-1}$$

6. asnaseer

For the final step, you can use the hints given to you by @experimentX

7. asnaseer

Please let me know if you require any further explanation.

8. PhoenixFire

I will attempt this in a few hours again... I have to go into Uni right now.

9. asnaseer

ok - good luck! :)

10. PhoenixFire

Okay I have 2 questions now. 1) Could I not state that from my Inductive Hypothesis (the statement is true for n=k) and let $B_1=(A_1A_2...A_k)$$B_2=A_{k+1}$ and then show: $(A_1A_2..A_nA_{n+1})^{-1}=B_2^{-1}B_1^{-1}$ From Ind. Hyp. with an n=2 $(A_1A_2..A_nA_{n+1})^{-1}=(B_1B_2)^{-1}$And then substitute in B1 and B2 again to result in an equality of Left hand side and Right hand side? 2) If I choose my base step as n=2 and want to prove $(A_1A_2)^{-1}=A_2^{-1}A_1^{-1}$Could I use a direct proof: Multiply both sides by A1A2$(A_1A_2)^{-1}A_1A_2=A_2^{-1}A_1^{-1}A_1A_2$$I=A_2^{-1}A_2$$I=I$Proving it for n=2. Am I correct to do it in this way?

11. experimentX

yeah you are correct ... you can use this law to show for n=3,4,... up to n

12. experimentX

for the last step, assume it to be true upto n, show that it is true for n+1 then this must be true for any n inductively.

13. PhoenixFire

Excellent. I understand it now. Out of curiosity, is my direct proof also valid?

14. experimentX

from this particular point, it's correct ... for n=2 2) If I choose my base step as n=2 and want to prove ------------

15. experimentX

now show it for n=3 and n=4

16. PhoenixFire

Well for 3 and 4 I'd use pretty much the same direct proof to reduce it to an Identity on both side proving an equality.

17. experimentX

yes you can do that ... let it assume to be true for some n. then show it to be true for n+1 ..... you can do this by assuming $$A_1 A_2 ...A_n = A$$ so that $$A_1 A_2 ...A_n A_{n+1} = A\times A_{n+1}$$ Now we assume $$(A_1 A_2 ...A_n)^{-1} = A_n^{-1} ...A_2^{-1}A_1^{-1}$$ we show that $$(A_1 A_2 ...A_nA_{n+1})^{-1} = A_{n_+1}^{-1}A_n^{-1} ...A_2^{-1}A_1^{-1}$$

18. experimentX

doing that will make your matrix simple product of two matrices. use the simple trick and assumption to show that property.

19. PhoenixFire

Great! Thank you for your help!

20. experimentX

yw.