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Author Topic: Infinite Solutions  (Read 4752 times)  Share 

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enwiabe

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Re: Infinite Solutions
« Reply #15 on: January 25, 2008, 02:39:54 pm »
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Yes, in what's called an augmented matrix (Mao, you'll come to it). You could get a system of equations:

1 2 3 | 0
2 2 2 | 2
4 4 4 | 4

The first 3/4 of the matrix represents x, y and z. And the number proceeding it represents the number it equals. I.E. x + 2y + 3z = 0 is represented by the first line 1 2 3 | 0

Now, in that 2nd row it's 2x + 2y + 2z = 2
In the 3rd row, 4x + 4y + 4z = 4

This matrix has a rank of TWO because when you row-reduce it (that is, perform elementary row operations in order to obtain leading ones), you may only obtain 2 leading ones. I will perform the row reductions below:

1 2 3 | 0
2 2 2 | 2
4 4 4 | 4

Row 1 is represented by 1 2 3 | 0, Row 2 is 2 2 2 | 1 and row 3 is 0 0 0 | 0

If you minus 2xR2 from R3 you get:

1 2 3 | 0
2 2 2 | 2
0 0 0 | 0

If you minus 2 x R1 from R2, you obtain the matrix:

1  2  3  | 0
0 -2 -4 | 2
0  0   0 | 0

Now divide R2 by -2 to obtain:

1 2 3 | 0
0 1 2 | -1
0 0 0 | 0

If you notice, we have two leading ones. The best explanation of this that I can give is a '1' in the left-hand most COLUMN, when ONLY preceded by a 0 or is itself in the first column. (A shitty definition, I know, maybe Ahmad can clear it up better).

Anyway from this you see that in a 3x3 matrix the rank is only TWO. The rank of a matrix is how many leading 1s it has. For there to be a distinct solution, we need Rank(M) >= Nj, where Nj = the number of columns in the matrix. However, for this matrix, Nj = 3 and Rank(M) = 2. Rank(M) < Nj, therefore we have more unknowns than we have equations to solve them! Thus, we have infinitely many solutions.

ALSO, this leads to a bit of a lemma. For any system of equations for which the number of equations is equal to the number of unknowns, if one equation is a scalar multiple of another, then the system of equations has INFINITELY many solutions.

« Last Edit: January 25, 2008, 02:49:10 pm by enwiabe »

midas_touch

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Re: Infinite Solutions
« Reply #16 on: January 25, 2008, 04:07:57 pm »
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I remember doing this in semester 1 uni last year. Enwiabe pretty much explained it right, If u have more non zero columns than non zero rows, then ur introducing a parameter, resulting in infinitely many solutions.
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phagist_

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Re: Infinite Solutions
« Reply #17 on: January 25, 2008, 05:03:35 pm »
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Put each coefficient in a matrix so its 3x3.

Get the determinant in terms of a and solve for zero.

You get a=0 or a=2.. then just look at the equations and using commonsense it's obvious that if a=0 the equations are not the same, hence there are not infinite solutions for that value of a.

Do the same with a=2.. then see that they are the same (I hope.. I haven't checked haha)

dcc

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Re: Infinite Solutions
« Reply #18 on: January 25, 2008, 05:19:21 pm »
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btw the stuff with the F-SWAP integral was a joke tbh

Ahmad

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Re: Infinite Solutions
« Reply #19 on: January 25, 2008, 05:44:02 pm »
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You get a=0 or a=2.. then just look at the equations and using commonsense it's obvious that if a=0 the equations are not the same, hence there are not infinite solutions for that value of a.

Do the same with a=2.. then see that they are the same (I hope.. I haven't checked haha)

Are you sure? :)
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/0

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Re: Infinite Solutions
« Reply #20 on: January 25, 2008, 05:53:45 pm »
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Thanks so far pplz, BTW I checked the answers and the answer is supposedly a = 0. Well, let's see what we get...







Whaddawhat? They don't look very much alike, do they  :(

And my calculator can only graph one 3D graph at a time, so I can't even check to see what they're like
« Last Edit: January 25, 2008, 06:02:42 pm by DivideBy0 »

Ahmad

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Re: Infinite Solutions
« Reply #21 on: January 25, 2008, 06:42:20 pm »
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They don't have to look alike. For there to be infinite solutions one equation must be redundant (i.e. add no more information than the other two). This means one equation is a linear combination of the other 2. This occurs for this system when a = 0. To check this note that the last equation is twice the first plus twice the second.

Note: When I saw adding equations, it means the obvious thing, that is, adding the corresponding left hand and right hand sides. :)
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phagist_

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Re: Infinite Solutions
« Reply #22 on: January 25, 2008, 06:48:46 pm »
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Are you sure? :)
nope haha, as I said I just got the two a values from solving the determinant as equal to zero. I just saw Mao say the answer was 2 so I didn't check, I just went with it.

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Re: Infinite Solutions
« Reply #23 on: January 25, 2008, 08:01:41 pm »
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They don't have to look alike. For there to be infinite solutions one equation must be redundant (i.e. add no more information than the other two). This means one equation is a linear combination of the other 2. This occurs for this system when a = 0. To check this note that the last equation is twice the first plus twice the second.

Note: When I saw adding equations, it means the obvious thing, that is, adding the corresponding left hand and right hand sides. :)

Thanks, what about the criteria for no solution?

phagist_

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Re: Infinite Solutions
« Reply #24 on: January 25, 2008, 08:30:37 pm »
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They'd (graphically) have to be parallel as either planes (in 3d) or lines (in 2d) I'd assume.

i.e one is a multiple of the other.

Collin Li

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Re: Infinite Solutions
« Reply #25 on: January 25, 2008, 10:01:48 pm »
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What are you guys exactly looking for?

Thanks, what about the criteria for no solution?

This occurs when you have the same equation twice with different solutions (i.e.: and )
« Last Edit: January 25, 2008, 10:04:04 pm by coblin »

Neobeo

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Re: Infinite Solutions
« Reply #26 on: January 26, 2008, 12:10:38 am »
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determinant =

The system has zero or infinite solutions iff determinant = 0.




From here you could probably do a rref on each solution to check whether it gives you a zero-solution system or an infinite-solution system.


The presence of {0,0,0,0} suggests that it has infinite solutions.



The presence of {0,0,0,1} suggests that it has no solutions.
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Ahmad

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Re: Infinite Solutions
« Reply #27 on: January 26, 2008, 09:11:51 am »
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They don't have to look alike. For there to be infinite solutions one equation must be redundant (i.e. add no more information than the other two). This means one equation is a linear combination of the other 2. This occurs for this system when a = 0. To check this note that the last equation is twice the first plus twice the second.

Note: When I saw adding equations, it means the obvious thing, that is, adding the corresponding left hand and right hand sides. :)

Thanks, what about the criteria for no solution?

The system must be inconsistent, in Neo's post he arrived at 0x + 0y + 0z = 1 which is impossible. You can also try imagining what's happening in terms of the planes in 3D, or you might think of the equations using a vector picture, where the x, y, z coefficients forms vectors, . Then there are no linear combinations of these vectors which add to , since

For example, this may happen if the 3 vectors span a plane (i.e. any linear combination of the 3 vectors results in a vector on a plane), and the right hand side vector is not on the plane. Note that if the RHS vector IS on the plane then you'd get infinite solutions. :)
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