IA (HL) Portfolio Type I -- Patterns Within Systems of Linear Equations

[Bishup]

for the 2x2 use letters to denote any number in your linear equation. Solve the linear equation using letters which will give you the same answer as you would get from solving any 2x2 that follows the same pattern as example given. In the end you will 2n=y and x=-n. N being any number so could be 1 so there y=2 and x=-1.

I actually had a solid formula and for this part of the IA I did not delve into matrices since it only involves a 2x2 but I suggest you have a look at matrices later on.

Can you help me prove the 2x2 in Section B and the 3x3 in Part A. I got ideas but can formulate a good proof.

Sorry I couldn't give the direct answer but I'm not allowed to.

Edited May 14, 2010 by sweetnsimple786
Please use the edit button to add more to a post when appropriate. :)

[alise]

In part A the 3x3 can be solved similar as 2x2. Just write it with letters and solve it somehow. F.e., Gauss' method..

About part B - I'm gonna get to it soon.

[The Jackal]

Gaussian method is not obligatory though, I did it using the addition method. The place where I am really stuck, is where I have to describe the significance of the solution.

Edited May 14, 2010 by The Jackal

[Leventhan]

for the 2x2 use letters to denote any number in your linear equation. Solve the linear equation using letters which will give you the same answer as you would get from solving any 2x2 that follows the same pattern as example given. In the end you will 2n=y and x=-n. N being any number so could be 1 so there y=2 and x=-1.

I actually had a solid formula and for this part of the IA I did not delve into matrices since it only involves a 2x2 but I suggest you have a look at matrices later on.

Can you help me prove the 2x2 in Section B and the 3x3 in Part A. I got ideas but can formulate a good proof.

Sorry I couldn't give the direct answer but I'm not allowed to.

ax + (a+d)y = a + 2d

bx + (b+d)y = b + 2d

Like this?

[Bishup]

ax + (a+d)y = a + 2d

bx + (b+d)y = b + 2d

Like this?

It is exactly that idea! I just used different letters.

[pnkct]

I have a question about part B. I have found out that all the equations written in the form ux+ury=ur^2 are tangents of curve x=-1/4y^2. Does the task have something to do with that, because I am not sure how to set up a general 2x2 system that incorporates this pattern.

Yes, I think this portfolio has something to do with that. For example in the first part (A) I found out that it's slope is -1. That means these two lines intersect at a point. I think thats why we have to display our solution graphically. I'm sure in part B we have to do something close to what you had said.

Can somebody help me with extending my investigation to 3x3 systems whose constants exhibithe pattern? I found the pattern to be c.x+(c+1)y = x+2 and dx+(d-3)y=d-6. I can't go any further than this (

Edited October 27, 2010 by Fizzy Dee

[Kaneele]

I have a question about part B. I have found out that all the equations written in the form ux+ury=ur^2 are tangents of curve x=-1/4y^2. Does the task have something to do with that, because I am not sure how to set up a general 2x2 system that incorporates this pattern.

Yes, I think this portfolio has something to do with that. For example in the first part (A) I found out that it's slope is -1. That means these two lines intersect at a point. I think thats why we have to display our solution graphically. I'm sure in part B we have to do something close to what you had said.

It's nice to use some programs, for example, Autograph or Graph - draw about 30 straight lines varying their constants (e.g. in one r=5 in other r=1/5 in the next one r=-5 in the next one r=-1/5). Then see how it turns out. Thanks to Mr Jackal for this advice

Edited May 15, 2010 by Kaneele

[Bishup]

Can anybody solve algebraically the proof for the 2x2 in part B.

I know the equation of the curve is y²=-4x

I know you're meant to have quadratic equation and rearrange it to get the above but can't remember how to do it HELP .

[Kaneele]

Can anybody solve algebraically the proof for the 2x2 in part B.

I know the equation of the curve is y²=-4x

I know you're meant to have quadratic equation and rearrange it to get the above but can't remember how to do it HELP .

Try writing a system of two general formulas, same as you did in the part A. There you used a d - difference, now you have to use r - ratio; the formula is different as well, of course, because part B has a different pattern. Try to somehow get values of x and y.

Once that's done - the rest is quite easy You'll be able to see what to do afterwards.

Edited May 16, 2010 by Kaneele

[Bishup]

I tried with ax+ary=ar². Can you tell me how you did it?

[Kaneele]

I tried with ax+ary=ar². Can you tell me how you did it?

Try making a system of two different equations (e.g. one would be ax+ary=ar² and the other one would be bx+bny=bn²). Try to make some transfigurations (probably using Gauss' method), until you come across x and y values, which seem like you could use them in further investigation.

[gedx2]

The two lines are perpendictular to each other. Hopes this helps

Quick question.

I am doing the same IA and I have a question about part A

second bullet where it says

Display your solution graphically. What is the significance of the solution?

SO what is the significance of the solution?

yeah, I don't know if I can just ignore this point. Because the significance can be seen only later when compared to similar systems. But there is no significance to the solution itself...

[gedx2]

Can somebody help me with extending my investigation to 3x3 systems whose constants exhibithe pattern? I found the pattern to be c.x+(c+1)y = x+2 and dx+(d-3)y=d-6. I can't go any further than this (

OK, try x+2y+3z=4 and 3x+y-z=-3

[Kaneele]

I'm quite sure everyone's already handed it in by now...

At least for us the day to hand it in was this monday, which is three days ago.

[Bishup]

I'm quite sure everyone's already handed it in by now...

At least for us the day to hand it in was this monday, which is three days ago.

Same actually

Moral of the story do it all in the 1st 2 days you get handed it.

I think we should have a thread on our modelling IA but with loads of cooperation between each other.

Sound like an idea?

[sweetnsimple786]

The same thread can be used in a few months for Nov folks & next year and so on. =)

And the threads are here for people to get help.

But they can be abused. Don't let others think for you. There's no reason. Y'all are smart kids. Think through it for yourself, and get and give help as needed.

[Kaneele]

I think this co-operating thing wouldn't be such a good idea because, God help us, if the examinators get all our works. I asked my maths teacher just 'theoretically' if I, for example, handed in same IA as somebody else in the world, could there be any chance that we would get in trouble.

Her answer was short. And positive.

So, I'd prefer if we do as we did. We sort of guide each other in the direction etc, but we don't all agree on one way to do things and we defintiely do not write the same. The idea is similar, but differently approached. That's the best way.

And thanks, it's very nice to be praised for cleverness now and then

[pnkct]

for the 2x2 use letters to denote any number in your linear equation. Solve the linear equation using letters which will give you the same answer as you would get from solving any 2x2 that follows the same pattern as example given. In the end you will 2n=y and x=-n. N being any number so could be 1 so there y=2 and x=-1.

I actually had a solid formula and for this part of the IA I did not delve into matrices since it only involves a 2x2 but I suggest you have a look at matrices later on.

Can you help me prove the 2x2 in Section B and the 3x3 in Part A. I got ideas but can formulate a good proof.

Sorry I couldn't give the direct answer but I'm not allowed to.

ax + (a+d)y = a + 2d

bx + (b+d)y = b + 2d

Like this?

what do the d's stand for? I'm having a difficulty trying to understand it.

[Kaneele]

The 'd' represents the difference from an arithmetic progression.

For instance, if you have an arithmetic progression - 1;4;7;10;13;16....bazillion - the d=3, because it can be calculated by subtracting one of the members of the progression from the one that follows it, in this case: 7-4=3 or 10-7=3 et cetera.

[pnkct]

The 'd' represents the difference from an arithmetic progression.

For instance, if you have an arithmetic progression - 1;4;7;10;13;16....bazillion - the d=3, because it can be calculated by subtracting one of the members of the progression from the one that follows it, in this case: 7-4=3 or 10-7=3 et cetera.

thank you well that was a embarrassing thing to ask. so this "pattern" has something to do with arithmetic sequences?