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## PhoenixFire 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 one year ago one year ago Edit Question Delete Cancel Submit

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1. PhoenixFire
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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.

• one year ago
2. UnkleRhaukus
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*

• one year ago
3. experimentX
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write |dw:1349805212895:dw|

• one year ago
4. experimentX
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|dw:1349805366358:dw|

• one year ago
5. asnaseer
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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: $$(A_1A_2...A_k A_{k+1})^{-1}=A_{k+1}^{-1}A_k^{-1}..A_2^{-1}A_1^{-1}$$

• one year ago
6. asnaseer
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For the final step, you can use the hints given to you by @experimentX

• one year ago
7. asnaseer
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Please let me know if you require any further explanation.

• one year ago
8. PhoenixFire
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I will attempt this in a few hours again... I have to go into Uni right now.

• one year ago
9. asnaseer
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ok - good luck! :)

• one year ago
10. PhoenixFire
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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?

• one year ago
11. experimentX
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yeah you are correct ... you can use this law to show for n=3,4,... up to n

• one year ago
12. experimentX
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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.

• one year ago
13. PhoenixFire
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Excellent. I understand it now. Out of curiosity, is my direct proof also valid?

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

• one year ago
15. experimentX
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now show it for n=3 and n=4

• one year ago
16. PhoenixFire
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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.

• one year ago
17. experimentX
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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 $$(A_1 A_2 ...A_nA_{n+1})^{-1} = A_{n_+1}^{-1}A_n^{-1} ...A_2^{-1}A_1^{-1}$$

• one year ago
18. experimentX
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doing that will make your matrix simple product of two matrices. use the simple trick and assumption to show that property.

• one year ago
19. PhoenixFire
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Great! Thank you for your help!

• one year ago
20. experimentX
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yw.

• one year ago
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