IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IB => Math => Topic started by: nid404 on January 01, 2010, 06:56:00 am

Title: Matrix Powers
Post by: nid404 on January 01, 2010, 06:56:00 am: Some1 please solve the last two questions
Thanks
Title: Re: Matrix Powers
Post by: eightAs on January 02, 2010, 11:13:29 am: $A = \begin{pmatrix} k + 1 & k - 1\\ k - 1 & k + 1 \end{pmatrix}$
$A^n = \2^{n-1} \begin{pmatrix} k^n + 1 & k^n - 1\\ k^n - 1 & k^n + 1 \end{pmatrix}$
Not sure about the proof yet. This is based only on pattern spotting.
Title: Re: Matrix Powers
Post by: eightAs on January 02, 2010, 11:34:04 am: Ok prove it using Mathematical Induction
We want to show that the statement is true for all $n \in \mathbb{Z^{+}}$
$ A^1 is\begin{pmatrix} k + 1 & k - 1\\ k - 1 & k + 1 \end{pmatrix}$ .
Now substitue values in the formula to check. Therefore the formula is true for n = 1.
We may assume that our formula holds for n. And multiply $A^n$ with $A$ to evaluate $A^{n+1}$ and simplify to show that the formula is true for (n + 1) as well. The calculations would look ugly even in Latex, so I am skipping them.
Title: Re: Matrix Powers
Post by: nid404 on January 02, 2010, 12:30:05 pm: thanks :)...actually im doin AS and don't a know a thing abt this portfolio....this is one of my friend's

Could u help me by telling how lengthy the response for the questions should be...as in the explanation
Title: Re: Matrix Powers
Post by: eightAs on January 02, 2010, 01:41:18 pm: Quote from: nid404 on January 02, 2010, 12:30:05 pm
thanks :)...actually im doin AS and don't a know a thing abt this portfolio....this is one of my friend's

Could u help me by telling how lengthy the response for the questions should be...as in the explanation
I'm not an IB student either. I do AS as well and the only reason I knew Mathematical induction is because it is there in the Further math syllabus. I'd rather not comment :-X
Title: Re: Matrix Powers
Post by: nid404 on January 02, 2010, 01:45:16 pm: Cuz we found the general form...the same thing...but we didn't know how to prove it as such....and describe it as they ask :-\

Anyway thanks a lot for your help
Title: Re: Matrix Powers
Post by: astarmathsandphysics on January 02, 2010, 03:56:22 pm: I will do it when I get home
Title: Re: Matrix Powers
Post by: eightAs on January 03, 2010, 03:27:36 am: Ok I'll put forward a general induction argument. This is going to be a long post. I have an IB HL book which lists out five steps involved in an induction proof. They are:
1. Prove the result for an initial value usually (n = 1).
2. Assume the result is true for another value n = k, k>1 stating the result.
3. Consider the case n = k + 1, writing down the goal clearly - the required form.
4. Using the assumption, now show that the result is true for n = k + 1.
5. Communicate why this proves this result using mathematical induction.
There are some comments at the end of the proof.

Our conjecture is $A^n = \2^{n-1} \begin{pmatrix} k^n + 1 & k^n - 1\\ k^n - 1 & k^n + 1 \end{pmatrix}$
where $A = \begin{pmatrix} k + 1 & k - 1\\ k - 1 & k + 1 \end{pmatrix}$
Write it as: $A^n = \begin{pmatrix} 2^{n-1}(k^n + 1) & 2^{n-1}(k^n - 1)\\ 2^{n-1}(k^n - 1) & 2^{n-1}(k^n + 1) \end{pmatrix}$

Step 1: when n = 1. $A^1 = \begin{pmatrix} 2^{1-1}(k^1 + 1) & 2^{1-1}(k^1 - 1)\\ 2^{1-1}(k^1 - 1) & 2^{1-1}(k^1 + 1) \end{pmatrix} = \begin{pmatrix} (k + 1) & (k - 1)\\ (k - 1) & (k + 1) \end{pmatrix}$ which is clearly true since A raised to 1 is A itself.

Step 2: So lets assume for some integer j, the formula is true. Notice we're not using n = k here (as described in the five steps above) as we already have a variable k.
$A^j = \2^{j-1} \begin{pmatrix} k^j + 1 & k^j - 1\\ k^j - 1 & k^j + 1 \end{pmatrix}$

Step 3: Substitute n = j + 1 in the original equation.
$A^{j+1} = \2^{j} \begin{pmatrix} k^{j+1} + 1 & k^{j+1} - 1\\ k^{j+1} - 1 & k^{j+1} + 1 \end{pmatrix} = \begin{pmatrix} (2^{j})(k^{j+1} + 1) & (2^{j})(k^{j+1} - 1)\\ (2^{j})(k^{j+1} - 1) & (2^{j})(k^{j+1} + 1) \end{pmatrix}$
This is what we have to prove, using the assumption in Step 2.

Step 4: The ugly calculations! $A^{j+1} = A^j.A = \begin{pmatrix} 2^{j-1}(k^j + 1) & 2^{j-1}(k^j - 1)\\ 2^{j-1}(k^j - 1) & 2^{j-1}(k^j + 1) \end{pmatrix} . \begin{pmatrix} (k + 1) & (k - 1)\\ (k - 1) & (k + 1) \end{pmatrix} = \begin{pmatrix} (k + 1)(k^j + 1)(2^{j-1}) + (k - 1)(k^j - 1)(2^{j-1}) & (k - 1)(k^j + 1)(2^{j-1}) + (k + 1)(k^j - 1)(2^{j - 1}) \\ (k - 1)(k^j + 1)(2^{j-1}) + (k + 1)(k^j - 1)(2^{j - 1}) & (k + 1)(k^j + 1)(2^{j-1}) + (k - 1)(k^j - 1)(2^{j-1}) \end{pmatrix} = \begin{pmatrix} 2^{j-1}(k^{j+1} + k^j + k + 1 + k^{j+1} - k^j - k + 1) & 2^{j-1}(k^{j+1} - k^j + k - 1 + k^{j+1} - k + k^j - 1) \\ 2^{j-1}(k^{j+1} - k^j + k - 1 + k^{j+1} - k + k^j - 1) & 2^{j-1}(k^{j+1} + k^j + k + 1 + k^{j+1} - k^j - k + 1) \end{pmatrix} = \begin{pmatrix} 2^{j-1}(2(k^{j+1} + 1)) & 2^{j-1}(2(k^{j+1} - 1)) \\ 2^{j-1}(2(k^{j+1} - 1)) & 2^{j-1}(2(k^{j+1} + 1)) \end{pmatrix} = \begin{pmatrix} (2^{j})(k^{j+1} + 1) & (2^{j})(k^{j+1} - 1)\\ (2^{j})(k^{j+1} - 1) & (2^{j})(k^{j+1} + 1) \end{pmatrix}$

Step 5: Conclude that this matrix is same as the one that appeared at the end of step 3. So the result is true for n = j + 1 when true for n = j. Since it is true for n = 1, it is true $\forall n \in Z^{+}$ by the principle of Mathematical Induction.

Since the project is being judged for presentation, notation and whatnot. I strongly recommend that your friend sticks to the notation I used, (as they are straight out of an IB (HL) book :P ) and lays out her proof in an organized manner. :)
Title: Re: Matrix Powers
Post by: nid404 on January 03, 2010, 05:37:31 am: omg! Thank you so very much for taking the pains to put this thing down...
THanks A Lot :D
Title: Re: Matrix Powers
Post by: eightAs on January 03, 2010, 05:38:27 am: Quote from: nid404 on January 03, 2010, 05:37:31 am
omg! Thank you so very much for taking the pains to put this thing down...
THanks A Lot :D
No problem. By the way, do you have the markscheme for this paper?
Title: Re: Matrix Powers
Post by: nid404 on January 03, 2010, 05:41:17 am: naah...it's not a paper...it's supposed to be their math portfolio for which they are marked...I have no idea about how it should look and my friend is unaware too cuz it is his first year in ib...

Anyway you have been of great help....And since your matter comes straight from IB HL I don't have to doubt it :P
Title: Re: Matrix Powers
Post by: sweet777 on January 30, 2010, 09:31:38 am: can u please help me solve this matrix question??

Write 5A²-6A=3I in the form AB=I and hence find A^-1 in terms of A and I

please help fasssssssssttttttttttttt!!!!!!!!!!!!!!!!!!!!!!!