Thanks, got the email. I understand why they have the X and Y now. It's so they can refer to the rotation matrix, (and reflection matrix) in the latter parts of the question.(Original post by Music99)
I'll give that a try, the X and Y thing is just weird.
PS: For part b), k isSpoiler:Show6

ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 21
 12042013 23:11

Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 22
 12042013 23:29
(Original post by ghostwalker)
Thanks, got the email. I understand why they have the X and Y now. It's so they can refer to the rotation matrix, (and reflection matrix) in the latter parts of the question.
PS: For part b), k isSpoiler:Show6 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 23
 13042013 14:43
(Original post by ghostwalker)
Thanks, got the email. I understand why they have the X and Y now. It's so they can refer to the rotation matrix, (and reflection matrix) in the latter parts of the question.
PS: For part b), k isSpoiler:Show6 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 24
 13042013 17:01
(Original post by Music99)
Hey, So I found that K=6 (as you did say in the spoiler ). Now I need to deduce the angle of rotation. I'm not too sure how to do this... Any hints?
1) You can apply the rotation matrix to a vector and then work out the angle between the resultant and the original.
Edit2: It actually needs a lot more work than that via this method  oops.
2) you know what 6 applications of the rotation matrix do.
Edit: Actually this last one is sufficient, as long as you bear in mind what 1 to 5 applications don't do.
The only bit that bugs me is how to differentiate between your angle and 360 minus your angle.Last edited by ghostwalker; 13042013 at 18:42. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 25
 13042013 18:33
(Original post by ghostwalker)
Two thoughts:
1) You can apply the rotation matrix to a vector and then work out the angle between the resultant and the original.
2) you know what 6 applications of the rotation matrix do.
Edit: Actually this last one is sufficient, as long as you bear in mind what 1 to 5 applications don't do.
The only bit that bugs me is how to differentiate between your angle and 360 minus your angle. 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 26
 13042013 18:43
(Original post by Music99)
But I'm still not sure how to go about getting the inital matrix, as it has to be a 3x3 rotation matrix. All I've got in my notes is that in 3d we rotate about a line, and that when we reflect we use half the angle.
Note: I corrected my previous post. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 27
 13042013 18:57
(Original post by ghostwalker)
I thought you were told it was a rotation matrix.
Note: I corrected my previous post.
So I've found the various equations for rotating about the x y and z axis, but as we aren't told what axis we are rotating around or anything I can't apply them.Last edited by Music99; 13042013 at 19:07. 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 28
 13042013 19:20
(Original post by Music99)
We are so we have the rotation matrix to the power of 6 is the 3x3 identity matrix, but then I don't see how from that I can get the angle of rotation. I'm so confused.
So, the power of 6 means that you've rotated through the same angle 6 times.
And that gives you the identity (matrix).
Do you see? 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 29
 13042013 19:25
(Original post by ghostwalker)
Well each multiplication by the rotation matrix rotates you through the same angle.
So, the power of 6 means that you've rotated through the same angle 6 times.
And that gives you the identity (matrix).
Do you see? 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 30
 13042013 19:34
(Original post by Music99)
so do you just do 2pi/6 which is pi/3. And that's the answer?
We need to consider the possibility that we haven't just rotated through 2pi, but rather through an integer multiple of 2pi.
E.g. 2(2p). Well if 6 applications of the rotation matrix take us through 4pi, then 3 applications would take us through 2pi, and so X^3 would be the identity matrix (where X is the rotation matrix). But X^3 is not the identity matrix so we can't have gone through 4pi.
We can eliminate all other multiples, except 5(2pi), which would give us 5pi/3 as the angle of rotation, or pi/3 in the opposite direction to that we had previously.
Edit: So, +/ pi/3.
Can't narrow it down any futher just by considering multiples of 2pi.
Is that good enough?
You could compare your matrix to the standard 3D rotation matrix  I think it's on wiki, if not in your notes.Last edited by ghostwalker; 13042013 at 20:00. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 31
 13042013 20:11
(Original post by ghostwalker)
That's one answer.
We need to consider the possibility that we haven't just rotated through 2pi, but rather through an integer multiple of 2pi.
E.g. 2(2p). Well if 6 applications of the rotation matrix take us through 4pi, then 3 applications would take us through 2pi, and so X^3 would be the identity matrix (where X is the rotation matrix). But X^3 is not the identity matrix so we can't have gone through 4pi.
We can eliminate all other multiples, except 5(2pi), which would give us 5pi/3 as the angle of rotation, or pi/3 in the opposite direction to that we had previously.
Edit: So, +/ pi/3.
Can't narrow it down any futher just by considering multiples of 2pi.
Is that good enough?
You could compare your matrix to the standard 3D rotation matrix  I think it's on wiki, if not in your notes.
Then it says by solving Xr=r and Yr=r where r is the column vector x,y,z or otherwise, find the axis of rotation for X and the plane of reflection for Y.
So we can rewrite the equations as (XI)r=0 , then I'm guessing we have to solve for the x,y,z and that gives us the axis of rotation. Not to sure about the second part. 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 32
 13042013 20:57
(Original post by Music99)
Ah okay, so we can right pi/3 pi/3 and 5pi/3.
Then it says by solving Xr=r and Yr=r where r is the column vector x,y,z or otherwise, find the axis of rotation for X and the plane of reflection for Y.
So we can rewrite the equations as (XI)r=0 , then I'm guessing we have to solve for the x,y,z and that gives us the axis of rotation. Not to sure about the second part.
In both cases you're finding the points which are not changed by the matrices.
In the case of the rotation matrix, this is the axis of rotation.
In the case of the reflection matrix, this is the plane of reflection. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 33
 13042013 21:42
(Original post by ghostwalker)
+/ pi/3 would cover it.
Yes.
In both cases you're finding the points which are not changed by the matrices.
In the case of the rotation matrix, this is the axis of rotation.
In the case of the reflection matrix, this is the plane of reflection.
so I subtracted the matrix and the identity to get
(1/3 2/3 1/3) (x)= (0)
( 1/3 1/3 2/3) (y)= (0)
(2/3 1/3 1/3 ) (z)= (0)
So then to find x,z,y I did A^1 times by 0,0,0, but that will just give 0,0,0 so is the axis of rotation the origin, or have I done something wrong? 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 34
 13042013 22:01
(Original post by Music99)
So I've tried to find the axis of rotation. I did (XI)r=0
so I subtracted the matrix and the identity to get
(1/3 2/3 1/3) (x)= (0)
( 1/3 1/3 2/3) (y)= (0)
(2/3 1/3 1/3 ) (z)= (0)
So then to find x,z,y I did A^1 times by 0,0,0, but that will just give 0,0,0 so is the axis of rotation the origin, or have I done something wrong?
Since the solution is going to be a line, not a unique point, the matrix is singular, and so you can't calculate an inverse.
You have 3 equations in 3 unknowns, but they are not independent.
You need to solve them to produce a line. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 35
 13042013 22:03
(Original post by ghostwalker)
In two dimension, the "axis" of rotation, is a point, but in three it's a line, so no, it's not (0,0,0), although that will of couse lie on the axis.
Since the solution is going to be a line, not a unique point, the matrix is singular, and so you can't calculate an inverse.
You have 3 equations in 3 unknowns, but they are not independent.
You need to solve them to produce a line. 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 36
 13042013 22:15
(Original post by Music99)
Yeah was thinking it's wrong, so i have to use gauss jordan elimination? 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 37
 13042013 22:22
(Original post by ghostwalker)
It's one method. I'm sure you're capable of solving what is now a standard matrix equation Ax=0 without having to check.Last edited by Music99; 13042013 at 22:33. 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 38
 13042013 23:08
(Original post by Music99)
Well the only solution I can see is x=0 y=0 z=0 for the axis of rotation for X and the plane of reflection of Y I tried inputting into a matrix solver and it says it's singular
Post your working for the axis of rotation one, and we'll see what's going on. 
Music99
 Follow
 1 follower
 2 badges
 Send a private message to Music99
 Thread Starter
Offline2ReputationRep: Follow
 39
 13042013 23:17
(Original post by ghostwalker)
Yes, they are singular, so you can't invert them.
Post your working for the axis of rotation one, and we'll see what's going on.
[1/3 2/3 1/3] [x]= 0
[1/3 2/3 2/3] [y]= 0
[2/3 1/3 2/3] [z]= 0
Then I just subbed that into a gauss jordan eliminator calculator online and it just comes up with x,y,z =0 
ghostwalker
 Follow
 83 followers
 13 badges
 Send a private message to ghostwalker
 Study Helper
Offline13Study Helper Follow
 40
 13042013 23:18
(Original post by Music99)
so we do (XI)r=0 where r is the column matrix x,y,z I is the identity and X is the rotation matrix, so we get
[1/3 2/3 1/3] [x]= 0
[1/3 2/3 2/3] [y]= 0
[2/3 1/3 2/3] [z]= 0
Then I just subbed that into a gauss jordan eliminator calculator online and it just comes up with x,y,z =0
Reply
Submit reply
Related discussions:
 matrix application question
 Im confused by this matrix transformation question (FP1)
 Matrix/Basis question
 FP4 Matrix Transformation Question??
 FP3, Further matrix algebra question.
 Singular matrix squared question
 3x3 Matrix rotation question?
 Matrix Transformation question
 Matrix Multiplication question help
 Unsure of what to do on a question (Matrix and general form)
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
Updated: April 20, 2013
Share this discussion:
Tweet