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# Matrix question Watch

1. (Original post by Music99)
I'll give that a try, the X and Y thing is just weird.
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 is
Spoiler:
Show
6
2. (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 is
Spoiler:
Show
6
Ah right makes sense I guess, kind of weirdly put in there though. Ah nice . Thanks I'll attempt the next bit tomorrow morning .
3. (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 is
Spoiler:
Show
6
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?
4. (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?
Two thoughts:

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.
5. (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.
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.
6. (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.
I thought you were told it was a rotation matrix.

Note: I corrected my previous post.
7. (Original post by ghostwalker)
I thought you were told it was a rotation matrix.

Note: I corrected my previous post.
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 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.
8. (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.
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?
9. (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?
so do you just do 2pi/6 which is pi/3. And that's the answer?
10. (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.
11. (Original post by ghostwalker)

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.
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 (X-I)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.
12. (Original post by Music99)
Ah okay, so we can right pi/3 -pi/3 and 5pi/3.
+/- pi/3 would cover it.

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 (X-I)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.
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.
13. (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've tried to find the axis of rotation. I did (X-I)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?
14. (Original post by Music99)
So I've tried to find the axis of rotation. I did (X-I)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?
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.
15. (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.
Yeah was thinking it's wrong, so i have to use gauss jordan elimination?
16. (Original post by Music99)
Yeah was thinking it's wrong, so i have to use gauss jordan elimination?
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.
17. (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.
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
18. (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
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.
19. (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.
so we do (X-I)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
20. (Original post by Music99)
so we do (X-I)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
Then, ditch the calculator, and do it by hand.

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