Linear Algebra Check...

Psiko · 3 Mar 2007 at 16:33

I'm fairly sure that some of your multiplication has gone wrong, in your answer for A^n. Do you realise that 11 x 1.1^n does not equal 12.1^n?

Also, do you understand that if (x^(i+1),y^(i+1)) = A (x^(i), y^(i)), then (x^(n), y^(n)) = A^n (x^(0), y^(0))?
You are told at the top of the question that x^(0) = 2000, y^(0) = 4000, so just plug this in to the equation above.

I hope that helps.

demon8991 · 12 Mar 2007 at 11:46

Psiko said:
I'm fairly sure that some of your multiplication has gone wrong, in your answer for A^n. Do you realise that 11 x 1.1^n does not equal 12.1^n?

Yer stupid mistake.

Psiko said:
Also, do you understand that if (x^(i+1),y^(i+1)) = A (x^(i), y^(i)), then (x^(n), y^(n)) = A^n (x^(0), y^(0))?
You are told at the top of the question that x^(0) = 2000, y^(0) = 4000, so just plug this in to the equation above.

Ok right ive sorted that out and got my A^n again, so all your saying i have to do is set n = 10 (for the number of days) work out that values for the matrix and then multiply it by the matrix:

(2000)
(4000)

The only reason why i ask is because in the question it says that i = number of days, but i cant see any other number that i can set n equal to.

Im hoping this is right lol!

FreeStream · 12 Mar 2007 at 11:53

Is this for A-Level maths?

I hope not lol!! :eek:

syke · 12 Mar 2007 at 11:56

/exits thread quickly.

~>Dg<~ · 12 Mar 2007 at 11:57

haaha good ol algebra our teacher was obsessed with it he used to rant and rave saying you dont know how impotant it is and set us unsolvable problems like four blackboards long fool.

demon8991 · 12 Mar 2007 at 12:01

Lol i think it might be in the furthur maths a level but i didnt do that.

Im doing it at uni.

I get my final answers as x = 9510.3885 (which i presume we round up to 9511) and y = 6052.0659112 (which i presume we round up to 6053)

Am i correct?

DaveF · 12 Mar 2007 at 12:27

demon8991 said:
Lol i think it might be in the furthur maths a level but i didnt do that.

Im doing it at uni.

I get my final answers as x = 9510.3885 (which i presume we round up to 9511) and y = 6052.0659112 (which i presume we round up to 6053)

Am i correct?

Sorry, missed this thread for a bit and don't really want to go through all the algebra again.

In practical terms, assuming (as you're on a computer forum) that you can program a computer in some form or other, it's easy enough to work out what x,y must be numerically. (Just keep multiplying by your original matrix).

Or, if you had to find A^10 by hand, note that to calculate A^10, you only need to calculate A^2, A^4,A^8 and then multiply A^2 and A^8, which is only 4 matrix multiplies.

The sad truth is that for n as small as 10, it's much easier to do this than muck about with eigenvectors and eigenvalues - so you could say this question is a complete waste of time...

demon8991 · 12 Mar 2007 at 12:45

Ahhh ok yer thats kool well ill check my answer that way.

Just interested to know does my working sound ok?

demon8991 said:
Ok right ive sorted that out and got my A^n again, so all your saying i have to do is set n = 10 (for the number of days) work out that values for the matrix and then multiply it by the matrix:

(2000)
(4000)

The only reason why i ask is because in the question it says that i = number of days, but i cant see any other number that i can set n equal to.

Im hoping this is right lol!

demon8991 · 12 Mar 2007 at 13:56

Ok ive prooved that my A^10 is correct which is encouraging.

It just strikes me odd that my x value after 10 days in larger than my y value, taking into consideration that intially my y value was bigger when we started, although i know its possible.

demon8991 · 12 Mar 2007 at 15:15

Ok well i think ive completed the bit of work but it all pivots on whether this theory is correct, as im not sure becuase its got a different number of columns to the matrix its being multiplied by.

Is this correct.

(A B) (X)
(C D) (Y)

equals

(AX+BY)
(CX + DY)

Where AX+BY is the X value and CX DY is the Y value?

demon8991 · 13 Mar 2007 at 12:45

Is that right?

alex2zulu · 13 Mar 2007 at 18:10

i get something like that looks like an electric shaver at the end :eek:

demon8991 · 14 Mar 2007 at 11:49

demon8991 said:
Ok well i think ive completed the bit of work but it all pivots on whether this theory is correct, as im not sure becuase its got a different number of columns to the matrix its being multiplied by.

Is this correct.

(A B) (X)
(C D) (Y)

equals

(AX+BY)
(CX + DY)

Where AX+BY is the X value and CX DY is the Y value?

Lol!

Well if anyone can just confirm my working would be great.