wbanner2001
Badges: 11
Rep:
?
#1
Report Thread starter 1 week ago
#1
1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?



Name:  04D4008B-CA3D-4F07-A09A-9C1EB3593F33.jpg.jpeg
Views: 4
Size:  31.9 KB
Name:  096778BB-2FC6-443E-8D8B-5E2984F646C3.jpg.jpeg
Views: 6
Size:  26.1 KB
Name:  33923F1C-AB3F-482A-A16C-4425DE7995AE.jpg.jpeg
Views: 5
Size:  18.1 KB
0
reply
wbanner2001
Badges: 11
Rep:
?
#2
Report Thread starter 1 week ago
#2
(Original post by wbanner2001)
1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?



Name:  04D4008B-CA3D-4F07-A09A-9C1EB3593F33.jpg.jpeg
Views: 4
Size:  31.9 KB
Name:  096778BB-2FC6-443E-8D8B-5E2984F646C3.jpg.jpeg
Views: 6
Size:  26.1 KB
Name:  33923F1C-AB3F-482A-A16C-4425DE7995AE.jpg.jpeg
Views: 5
Size:  18.1 KB
Bump
0
reply
ghostwalker
  • Study Helper
Badges: 17
#3
Report 1 week ago
#3
(Original post by wbanner2001)
1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?
9c. You need to get it in the form x+iy before you can isolate the real part like that. So, mupltiply top & bottom by the complex conjugate of the denominator.
0
reply
Silurianwarrior1
Badges: 3
Rep:
?
#4
Report 1 week ago
#4
Name:  image.jpg
Views: 7
Size:  90.0 KB
I think this is correct?? Let me know.
It’s just playing around with the matrix identities.
So essentially M^-1 is equal to the transposed Matrix

Hence why only 1 Mark. I have absolutely no idea if that correct though...
0
reply
wbanner2001
Badges: 11
Rep:
?
#5
Report Thread starter 1 week ago
#5
(Original post by Silurianwarrior1)
Name:  image.jpg
Views: 7
Size:  90.0 KB
I think this is correct?? Let me know.
It’s just playing around with the matrix identities.
So essentially M^-1 is equal to the transposed Matrix

Hence why only 1 Mark. I have absolutely no idea if that correct though...
I got M^-1 = 1/18 M^T in the end for part c as replacing I with MM^-1 gives:
MM^T = 18MM^-1
M^T = 18 M^-1
M^-1 = 1/18 M^T

I am confusing myself over d now though aha. Thanks for the response 🙏
0
reply
RDKGames
Badges: 20
Rep:
?
#6
Report 1 week ago
#6
(Original post by Silurianwarrior1)
I think this is correct?? Let me know.
It’s just playing around with the matrix identities.
So essentially M^-1 is equal to the transposed Matrix

Hence why only 1 Mark. I have absolutely no idea if that correct though...
Not correct. You cannot cancel matrices like that generally, so don't get into that habit, although it's premitted here since this matrix is I.

(Original post by wbanner2001)
1) How do you do 8c quickly if it is 1 mark and consequently 8d?

2) Why is my answer to 9c nothing like the show that?
The fact that MM^T = kI means that M,M^T are pretty much inverses of each other. Not exactly though, because if they were then MM^T = I where k=1 but you already correctly determined it is k=18, so we would obtain a contradiction in saying that.

Just take their fact and divide by k then it's clear that

\dfrac{1}{k}MM^T = I

so obviously you now have M being multiplied by \frac{1}{k}M^T which outputs the identity matrix... not much to say here other than realise that this necessarily means M^{-1} = \frac{1}{k}M^T.
0
reply
wbanner2001
Badges: 11
Rep:
?
#7
Report Thread starter 1 week ago
#7
(Original post by ghostwalker)
9c. You need to get it in the form x+iy before you can isolate the real part like that. So, mupltiply top & bottom by the complex conjugate of the denominator.
I realised the denominator without converting it into x+iy form first and that worked - thank you!
0
reply
wbanner2001
Badges: 11
Rep:
?
#8
Report Thread starter 1 week ago
#8
For 9d I did this but got sin 2x = -4 which
is obviously invalid so can you see what I did wrong?


(Original post by ghostwalker7842412)
9c. You need to get it in the form x+iy before you can isolate the real part like that. So, mupltiply top & bottom by the complex conjugate of the denominator.
Name:  463188C2-6941-4C25-8DDE-6A4A98E5AFC1.jpg.jpeg
Views: 6
Size:  24.1 KB
0
reply
Silurianwarrior1
Badges: 3
Rep:
?
#9
Report 1 week ago
#9
(Original post by RDKGames)
Not correct. You cannot cancel matrices like that generally, so don't get into that habit, although it's premitted here since this matrix is I.



The fact that MM^T = kI means that M,M^T are pretty much inverses of each other. Not exactly though, because if they were then MM^T = I where k=1 but you already correctly determined it is k=18, so we would obtain a contradiction in saying that.

Just take their fact and divide by k then it's clear that

\dfrac{1}{k}MM^T = I

so obviously you now have M being multiplied by \frac{1}{k}M^T which outputs the identity matrix... not much to say here other than realise that this necessarily means M^{-1} = \frac{1}{k}M^T.
You CAN cancel IDENTITY matrices...
Hence why I cancelled them out. My answer of M^T = K(M^-1) is what you achieved also...
0
reply
RDKGames
Badges: 20
Rep:
?
#10
Report 1 week ago
#10
(Original post by Silurianwarrior1)
You CAN cancel IDENTITY matrices...
Hence why I cancelled them out. My answer of M^T = K(M^-1) is what you achieved also...
That's what I effectively said, but more importantly I told you to not get into the habit of cancelling out matrices between both sides. You cannot divide by a matrix and multiplication is not commutative.

ABC = DBE does not mean AC = DE.

You should take these properties more seriously by instead of putting a slash through I, instead note that I is such a matrix so that AI=IA=A. Also, since we don't have commutativity, you need to pre-multiply or post-multiply by M^{-1}. Hence the correct practice to get into for matrices is as follows;

MM^T = kI

M^{-1}MM^T = kM^{-1}I

IM^T = kM^{-1}

M^T = kM^{-1}


where at no point have I 'cancelled' out common matrices between both sides, and I have strictly pre-multiplied both sides by M^{-1}.
0
reply
ghostwalker
  • Study Helper
Badges: 17
#11
Report 1 week ago
#11
(Original post by wbanner2001)
For 9d I did this but got sin 2x = -4 which
is obviously invalid so can you see what I did wrong?
I would suggest checking the formula you have for S - you've not posted its derivation, but I'd guess that's where the problem lies.
0
reply
wbanner2001
Badges: 11
Rep:
?
#12
Report Thread starter 1 week ago
#12
(Original post by RDKGames)
That's what I effectively said, but more importantly I told you to not get into the habit of cancelling out matrices between both sides. You cannot divide by a matrix and multiplication is not commutative.

ABC = DBE does not mean AC = DE.

You should take these properties more seriously by instead of putting a slash through I, instead note that I is such a matrix so that AI=IA=A. Also, since we don't have commutativity, you need to pre-multiply or post-multiply by M^{-1}. Hence the correct practice to get into for matrices is as follows;

MM^T = kI

M^{-1}MM^T = kM^{-1}I

IM^T = kM^{-1}

M^T = kM^{-1}


where at no point have I 'cancelled' out common matrices between both sides, and I have strictly pre-multiplied both sides by M^{-1}.
Can you see what I need to do for 9d?
0
reply
wbanner2001
Badges: 11
Rep:
?
#13
Report Thread starter 1 week ago
#13
(Original post by ghostwalker)
I would suggest checking the formula you have for S - you've not posted its derivation, but I'd guess that's where the problem lies.
So I got C+iS in realised form as (16 -4 cos 2x + 4i sin 2x)/(17 -8 cos 2x + 8i sin 2x) and therefore S as (16+4sin 2x)/(17+8sin2x) before equating to 0. Is that right?
0
reply
RDKGames
Badges: 20
Rep:
?
#14
Report 1 week ago
#14
(Original post by wbanner2001)
Can you see what I need to do for 9d?

Don't have time to look into this too much at the moment, but for 8d have you tried solving the following system?

q+4r-s=0

3q + ps = 0

q^2 + r^2 + s^2 = 18
0
reply
wbanner2001
Badges: 11
Rep:
?
#15
Report Thread starter 1 week ago
#15
(Original post by RDKGames)
Don't have time to look into this too much at the moment, but for 8d have you tried solving the following system?

q+4r-s=0

3q + ps = 0

q^2 + r^2 + s^2 = 18
I had set up those three but the squares put me off... I will try again! Thanks
0
reply
RDKGames
Badges: 20
Rep:
?
#16
Report 1 week ago
#16
(Original post by wbanner2001)
I had set up those three but the squares put me off... I will try again! Thanks
Just sub things into that last equation so that you don't get into the motion of rooting things and making everything more complicated than it needs to be.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Regarding Ofqual's most recent update, do you think you will be given a fair grade this summer?

Yes (306)
34.27%
No (587)
65.73%

Watched Threads

View All