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Maths Help

Blogs > p23s3
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Play
Profile Blog Joined November 2007
Australia608 Posts
Last Edited: 2008-08-16 03:19:01
August 16 2008 02:52 GMT
#1
Hey guys

I've got a maths assignment on the go and there a few little things i need help with. I've searched everywhere and my tutor couldn't help me either, nor have my friends got any clue. I've done most of the assignment but a couple of things i dont know.

The assignment is on 2x2 Matrix transformations.

Heres a simple question that i cannot work out:

For the matrix M =

(-1 2)
(-2 3)

Find the determinant: Ok, its 1.

What does this tell you about the transformation represented by M?

I have no clue, i have looked around and have no idea what the determinant has to do with the transformation.

It then goes on to talk about invariants , and i'm cool with all that, but does anybody have any idea about what the determinant tells you about the matrix transformation?

The invariant points through transformation of Matrix M are (x, y) where x = y. ie (3,3). If that helps.

Hope that makes sense even though its kinda wack ><

edit: graph of random images after transformation

[image loading]


jmascis
Deleted User 3420
Profile Blog Joined May 2003
24492 Posts
August 16 2008 02:54 GMT
#2
the blue pill
Play
Profile Blog Joined November 2007
Australia608 Posts
August 16 2008 02:55 GMT
#3
On August 16 2008 11:54 travis wrote:
the blue pill


uh oh
jmascis
crabapple
Profile Blog Joined June 2008
United States397 Posts
August 16 2008 03:05 GMT
#4
sry i cna't give u an answer, but im pretty sure stuff like this is stated explicitly in some chapter's instructions. usually they will have a compilation of properties. cause matrices and linear algebra is basically a long chain of, "this means this which also means this which is the same as saying 5 other things"

my guess would be that the transform is reversible.
ShOoTiNg_SpElLs
Profile Joined July 2003
Korea (South)690 Posts
Last Edited: 2008-08-16 03:15:49
August 16 2008 03:08 GMT
#5
Well since the determinant isn't 0, you know the matrix isn't singular. So then the transformation represented by M is invertible, i.e. it is one to one and onto.
Play
Profile Blog Joined November 2007
Australia608 Posts
August 16 2008 03:24 GMT
#6
On August 16 2008 12:08 ShOoTiNg_SpElLs wrote:
Well since the determinant isn't 0, you know the matrix isn't singular. So then the transformation represented by M is invertible, i.e. it is one to one and onto.


so basically...its just going to transform points from (x,y) to (x', y')...?
jmascis
ShOoTiNg_SpElLs
Profile Joined July 2003
Korea (South)690 Posts
August 16 2008 03:33 GMT
#7
Each point (x, y) is mapped to a unique point (x', y'), yes (and there exists a transformation which maps (x', y') back to (x, y)).
BottleAbuser
Profile Blog Joined December 2007
Korea (South)1888 Posts
August 16 2008 03:37 GMT
#8
A "one to one" relationship means that for every point BEFORE being transformed only goes to one unique point AFTER being transformed, and also that every point that has already been transformed can only get there by transforming a unique starting point.

Onto means that for every point in the codomain (of points) that the function maps to, there is also a point in the domain (starting point) that will yield that after-transformed point.

-_- forgot all the correct terminology
Compilers are like boyfriends, you miss a period and they go crazy on you.
Hittegods
Profile Joined April 2007
Stockholm4641 Posts
August 16 2008 04:02 GMT
#9
Take the ring off.
This neo violence, pure self defiance
overpool
Profile Joined April 2008
United States191 Posts
Last Edited: 2008-08-16 04:56:46
August 16 2008 04:40 GMT
#10
From Wiki
Determinants are used to characterize invertible matrices (i.e., exactly those matrices with non-zero determinants)


So basically, since the determinant != 0, there is an "inverse matrix" that can "undo" any transformation. That's the best I can come up with...
yay i love tl events
wesbrown
Profile Joined March 2008
United States31 Posts
Last Edited: 2008-08-16 05:35:02
August 16 2008 05:22 GMT
#11
Suppose the matrix M represents the linear transformation T:V->V, and suppose S is some object in V with hyper-volume A. Then if you apply the linear transformation T to S by T(S)=S', S' has hyper-volume |det(M)|*A. (A negative determinant means the orientation of S is reversed.)

So, in the 2x2 case, if the determinant of M is 1, the linear transformation preserves the orientation and area of all objects in R^2 (or whatever vector space you're working in). For example, the unit square having coordinates (0,0), (0,1), (1,0), and (1,1) is transformed into a quadrilateral with coordinates (0,0), (2,3), (-1,-2), and (1,1) using the matrix you gave. It's easy to check the the area of the quadrilateral is 1, and the orientation is preserved as the point (1,0) is "pulled" to (-1,-2) while the point (0,1) is pulled to (2,3).

If you don't really get the orientation-reversing part, look at the matrix M'=((1,2),(2,3)), which has determinant -1. Transform the unit square using both M and M' and note the location of the corner points. (You already know them under M. For M', they are (0,0), (2,3), (1,2), and (3,5), in the same order as before.) If you think about the movement of the points under M and M', you should notice that (0,1) and (1,0) occupy different relative positions after the transformations. I apologize if this isn't very clear, but you should get it eventually if you draw the pictures.

Edit: I spent 2 minutes in Paint. Notice how the red and blue dots have switched sides with the det=-1 transformation.

[image loading]
Grobyc
Profile Blog Joined June 2008
Canada18410 Posts
August 16 2008 06:05 GMT
#12
holy shit im glad i dropped math for next year T_T
If you watch Godzilla backwards it's about a benevolent lizard who helps rebuild a city and then moonwalks into the ocean.
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