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# matrices -proof by induction watch

1. I have this question, which i seem to be able to do the most part of, but can't seem to figure out how to complete it..

Poove, by induction, [1, 1, 0, 1] ^x = [1, x, 0, 1] for all values of x.

Where the matrix is a 2x2 and reads across such that [A, B, C, D]

thanks =)
2. For your induction hypothesis assume [1, 1, 0, 1] ^k = [1, k, 0, 1] for some value of k.

Consider [1, 1, 0, 1] ^(k+1) = something X something...

Can you take it from there?
3. ive gotten that far...it's atually going through the inductive part that i get stuck with...i can't seem to actually proove it. But i can tell that im close..there is something that im just not doing right, and i can't work out what...

but thank you
4. I was doing this at college today! you say 'let it be true for x = k, and then write it the same but with k instead of x. Then you say, show to be true for x = k +1; which when you write it, is the matrice you wrote with k (matrice ^k = [1 k 0 1]) *times* the original matrice Because: something to the power of ^k+1 is the same as something to the power of k *times* soemthing to the power of one (just the original number/matrice) (rules of indices) - then you just multiply them and should be able to factor k+1 where x was at the start, hence showing it's true for k and K+1 therefore all values of x

hope at least some part of that made any sense...!
5. (Original post by macccj)
I was doing this at college today! you say 'let it be true for x = k, and then write it the same but with k instead of x. Then you say, show to be true for x = k +1; which when you write it, is the matrice you wrote with k (matrice ^k = [1 k 0 1]) *times* the original matrice Because: something to the power of ^k+1 is the same as something to the power of k *times* soemthing to the power of one (just the original number/matrice) (rules of indices) - then you just multiply them and should be able to factor k+1 where x was at the start, hence showing it's true for k and K+1 therefore all values of x

hope at least some part of that made any sense...!
yeah that makes loads of sense...thank you for that!! although, i might be being stupid...but then dont i get: [1, k, 0, 1] + [1, k+1, 0, 1] which doesnt get me the right answer?

unless i've confused myself and added something extra?
6. how come you're adding? if you end up with [ 1, k+1, 0, 1] after multiplying your matrices', you've hence shown that it's true... i think :L
7. hold up - you multiply [1, 1, 0, 1]^k by [ 1, 1, 0, 1]

which means you multiply [ 1, k, 0, 1] (because you said at the start let it be true that [1, 1, 0, 1]^k = [1, k, 0,1]) by [1, 1, 0, 1]
which should work out as [1, k+1, 0, 1]?
8. (Original post by macccj)
hold up - you multiply [1, 1, 0, 1]^k by [ 1, 1, 0, 1]

which means you multiply [ 1, k, 0, 1] (because you said at the start let it be true that [1, 1, 0, 1]^k = [1, k, 0,1]) by [1, 1, 0, 1]
which should work out as [1, k+1, 0, 1]?
yeah, you're right....that makes sense now... I had confused myslef by looking at the wrong thing when i was working it out jsut now...Thank you for your help! It is much appreciated!
9. no worries, glad you got it!
10. i have another one for you today...proof by induction again..

3^n > n^3 when n >3 ???

am i right by showing it's true for n=4 as that is greater than 3?

and then writing: assume true for n=k such that 3^k > k^3

i've then gone on to say that the next term is 3^k+1 therefore,: 3^k + 3^k+1 > k^3 + 3^k+1 ??

thanks
11. Yes you are right by showing it for .

You need to show that , given the assumption that . What you've written doesn't show that.
12. (Original post by Dragon)
Yes you are right by showing it for .

You need to show that , given the assumption that . What you've written doesn't show that.

thats what i was thinking..its what confused me..so how do i go about showing that im guessing where i was going wrong was thinking that i added them ut i dont need to do i...i just have to prove ?

thank you for helping
13. Expanding both sides is a good start.
14. yeah, when you multiply it out you get multiples of the given formula from the start, sorry i was no more help!!
15. thank you =)

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