You are Here: Home >< Maths

# Proof by induction(matrix version) Watch

1. http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
2. (Original post by huiop)
http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
I might be wrong, but I have a feeling the kth term is the determinant...
3. (Original post by huiop)
http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
But can you not do it for K+1, and use the result from the LHS of the equation in your assumption as a substitute?
4. (Original post by huiop)
http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
Test for n=1
Assume true for n=k.
Prove for n=k+1. This would require you you to post-multiply both sides by the matrix 1, -1, 0, 5. (You know which one, cba to post it in the form) and factor every element to show it works for n=k+1.
5. (Original post by huiop)
http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
Consider the matrix to the power of k + 1,and split that into and sub that kth power matrix for the general formula we assumed was true for n = k.
6. (Original post by huiop)
http://files.physicsandmathstutor.co...%20Edexcel.pdf
n=1

Assume true for n=k that

what am i supposed to do next? it doesn't give me something to substitute something somewhere?
Consider multiplying the RHS by on the right. Can you see why?
7. (Original post by Cadherin)
I might be wrong, but I have a feeling the kth term is the determinant...
Not needed for this proof. Though the determinant is for all ;
8. (Original post by Slowbro93)
But can you not do it for K+1, and use the result from the LHS of the equation in your assumption as a substitute?
(Original post by RDKGames)
Test for n=1
Assume true for n=k.
Prove for n=k+1. This would require you you to post-multiply both sides by the matrix 1, -1, 0, 5. (You know which one, cba to post it in the form) and factor every element to show it works for n=k+1.
u sure i can do that? i considered it but i wasn't too sure if i was allowed to do that since it didn't state that matrix without to the power of n on it
(Original post by AMarques)
Consider the matrix to the power of k + 1,and split that into and sub that kth power matrix for the general formula we assumed was true for n = k.

(Original post by Paraphilos)
Consider multiplying the RHS by on the right. Can you see why?
so you get the identity matrix on the LHS?
Edit:^^ no u don't but u get 1 of tht matrix multiplied by the kth matrix which is to the power k+1 right?
9. (Original post by huiop)
u sure i can do that? i considered it but i wasn't too sure if i was allowed to do that since it didn't state that matrix without to the power of n on it
Of course. Just like with summation of series when you add the (k+1)th term onto both sides, you multiply here by the same matrix as you're working with exponents.

Let

Then it follows that (think of it as performing a transformation A a total amount of k times, then doing an additional one)

And you know what is because you assumed it to be true for
10. (Original post by RDKGames)
Of course. Just like with summation of series when you add the (k+1)th term onto both sides, you multiply here by the same matrix as you're working with exponents.

Let

Then it follows that (think of it as performing a transformation A a total amount of k times, then doing an additional one)

And you know what is because you assumed it to be true for
i've got

the

but the other meaty term i've only gotten this far

11. (Original post by huiop)
i've got

the

but the other meaty term i've only gotten this far

Is it really meaty though?
12. (Original post by RDKGames)
Is it really meaty though?
Looks like we have another student (in the form of Mr or Mrs Huiop) who doesn't know their laws of indices. Typical.
13. (Original post by HapaxOromenon3)
Looks like we have another student (in the form of Mr or Mrs Huiop) who doesn't know their laws of indices. Typical.
Those are common around here.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: August 17, 2016
Today on TSR

### Last-minute PS help

100s of personal statements examples here

### Loneliness at uni

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.