Post by: /0 on January 25, 2008, 07:35:19 am
How can I do this? These parameters chapters seems pretty obscure to me. :idiot2: If it helps you can use matrices. Thanks
Post by: cara.mel on January 25, 2008, 09:27:09 am
I also remember that you may need to actually know it for a multi choice question x_x
Post by: Mao on January 25, 2008, 11:54:48 am
but if you have a TI-89T, then use rref(matrix), u should know that right??
yields the result
note: you can arrive at this with some pretty hectic substitution and those alike...
and then i'm lost...
Post by: cara.mel on January 25, 2008, 12:19:45 pm
This question is really annoying me, because I know I used to be able to do it. I can go find my methods stuff and look for it, but I think my first post has the general idea. That is same line = infinite solutions, random paralell lines = no solutions, anything else = 1 solution. Except that is all for 2 equations, I don't remember what to do with 3
Edit, yeah ignore me, that is all for 2 equations. *pissed off at this* =)
I even found my photocopied chapter from the methods cas book (we had to live with the normal methods book, go guinea-pig-ness)
Post by: dcc on January 25, 2008, 12:35:31 pm
(I should of done CAS)
edit: btw I tried those 'hectic substitutions' but I'm not sure how to figure out if a line in three dimensions is parallel or even how to figure out the graident of a line in three dimensions (imagine like m = rise / run / SIDEWAYS WTF LOL)
Post by: Mao on January 25, 2008, 12:38:24 pm
I don't know that Mao =P
This question is really annoying me, because I know I used to be able to do it. I can go find my methods stuff and look for it, but I think my first post has the general idea. That is same line = infinite solutions, random paralell lines = no solutions, anything else = 1 solution. Except that is all for 2 equations, I don't remember what to do with 3
same here... :(
there is a way with n variables.... but i really forgot how to do it
CAS is for mathematicians who cant add in their head.i thought in 3D they were planes.... ?? and intersections of planes are curves (hence infinite solution)
(I should of done CAS)
edit: btw I tried those 'hectic substitutions' but I'm not sure how to figure out if a line in three dimensions is parallel
and then they'd have points touching (turning points)
edit: oh wait, that's 2 relations in 3D, 3 relations should be SHITLOADS more complicated... :(
Post by: dcc on January 25, 2008, 12:39:12 pm
Post by: dcc on January 25, 2008, 12:39:55 pm
I don't know that Mao =P
This question is really annoying me, because I know I used to be able to do it. I can go find my methods stuff and look for it, but I think my first post has the general idea. That is same line = infinite solutions, random paralell lines = no solutions, anything else = 1 solution. Except that is all for 2 equations, I don't remember what to do with 3
same here... :(
there is a way with n variables.... but i really forgot how to do itCAS is for mathematicians who cant add in their head.i thought in 3D they were planes.... ??
(I should of done CAS)
edit: btw I tried those 'hectic substitutions' but I'm not sure how to figure out if a line in three dimensions is parallel
well im assuming the x-y-z like i-j-k in spec
Post by: cara.mel on January 25, 2008, 12:45:40 pm
man seeing this, im glad I didn't do CAS, seems tedious :P (harder to get good scores as well)
I really didn't have a choice. Year 11 methods class turned into guinea pigs ftl :(
To solve the question, the only thing I have thought of is trying to get 2 equations that are the same. Otherwise, a=2 could be possible. I really dont know *headdesks*
Post by: dcc on January 25, 2008, 12:52:00 pm
man seeing this, im glad I didn't do CAS, seems tedious :P (harder to get good scores as well)
I really didn't have a choice. Year 11 methods class turned into guinea pigs ftl :(
To solve the question, the only thing I have thought of is trying to get 2 equations that are the same. Otherwise, a=2 could be possible. I really dont know *headdesks*
:( Guinea Pigs
well.
from what Mao got from his calculator, i assume that means:
And its obvious that:
so those two solutions are parallel? lol
Post by: Mao on January 25, 2008, 01:39:11 pm
i was procrastinating and i got the solution by hand now as well :P
but it doesnt get anywhere....
btw wouldnt the final matrix need to be:
Post by: Mao on January 25, 2008, 01:50:52 pm
solving for the intersections between the equations:
PS: i know the answer is
Post by: dcc on January 25, 2008, 01:52:26 pm
Now, taking the F-integral-SWAP of that matrix, we arrive at:
QED
Post by: cara.mel on January 25, 2008, 02:05:24 pm
Mao: I thought the special matrixes were like
1 0 0
0 1 0
0 0 1
No idea what that means any more xD
Your wild attempts are about where I got up to before I gave up
Post by: Mao on January 25, 2008, 02:29:07 pm
dcc, I have no idea what that means :Pyeah, but that's for distinct solutions
Mao: I thought the special matrixes were like
1 0 0
0 1 0
0 0 1
No idea what that means any more xD
Your wild attempts are about where I got up to before I gave up
for infinite solution i thought there has to be a row of "0 0 0" and such?
Post by: enwiabe on January 25, 2008, 02:39:54 pm
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.
Post by: midas_touch on January 25, 2008, 04:07:57 pm
Post by: phagist_ on January 25, 2008, 05:03:35 pm
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)
Post by: dcc on January 25, 2008, 05:19:21 pm
Post by: Ahmad on January 25, 2008, 05:44:02 pm
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? :)
Post by: /0 on January 25, 2008, 05:53:45 pm
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
Post by: Ahmad on January 25, 2008, 06:42:20 pm
Note: When I saw adding equations, it means the obvious thing, that is, adding the corresponding left hand and right hand sides. :)
Post by: phagist_ on January 25, 2008, 06:48:46 pm
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.
Post by: /0 on January 25, 2008, 08:01:41 pm
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?
Post by: phagist_ on January 25, 2008, 08:30:37 pm
i.e one is a multiple of the other.
Post by: Collin Li on January 25, 2008, 10:01:48 pm
Thanks, what about the criteria for no solution?
This occurs when you have the same equation twice with different solutions (i.e.:
Post by: Neobeo on January 26, 2008, 12:10:38 am
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.
Post by: Ahmad on January 26, 2008, 09:11:51 am
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,
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. :)