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

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- PhoenixFire

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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.

- UnkleRhaukus

*

- experimentX

write |dw:1349805212895:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- experimentX

|dw:1349805366358:dw|

- 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: \((A_1A_2...A_k A_{k+1})^{-1}=A_{k+1}^{-1}A_k^{-1}..A_2^{-1}A_1^{-1}\)

- asnaseer

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

- asnaseer

Please let me know if you require any further explanation.

- PhoenixFire

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

- asnaseer

ok - good luck! :)

- 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?

- experimentX

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

- 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.

- PhoenixFire

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

- 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
------------

- experimentX

now show it for n=3 and n=4

- 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.

- 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
\( (A_1 A_2 ...A_nA_{n+1})^{-1} = A_{n_+1}^{-1}A_n^{-1} ...A_2^{-1}A_1^{-1} \)

- experimentX

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

- PhoenixFire

Great! Thank you for your help!

- experimentX

yw.

Looking for something else?

Not the answer you are looking for? Search for more explanations.