Matrix problem Watch

bigmansouf
Badges: 20
Rep:
?
#1
Report Thread starter 1 week ago
#1
Question:
Matrices P and Q are members of a set R which is defined as follows:
 R= \left \{ \begin{pmatrix} a &b \\  c &d  \end{pmatrix}: a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}
Prove that the product PQ is also a member of set R.

Solution
let
 P=\begin{pmatrix} l &k \\  k & l \end{pmatrix}
 Q = \begin{pmatrix} u & v \\ u & -v \end{pmatrix}

QP=\begin{pmatrix} (lu+kv) & (lv-ku) \\ (ku+lv)&(kv-lu)\end{pmatrix}
since  ad-bc=1
({kv+lu)(kv-lu)}{(ku+lv)(lv-ku)=1}
(kv^2-lu^2)-(lv^2-ku^2)=1
(v^2+u^2)(k-l)=1
(v^2+u^2)=k-l
this is the part i am stuck at
please help me
Last edited by bigmansouf; 5 days ago
0
reply
begbie68
Badges: 16
Rep:
?
#2
Report 1 week ago
#2
What else was there in the question?
... you've defined P&Q to be particular cases of R. Was this part of the question?
(eg the use of a,b,c,d in the set of R matrices would suggest that a,b,c,d are different values, whereas you've used matrices for P&Q which have repeated values for some elements.)
Do you need to find relationships between l,k and u,v ?

From what you've shown there, we have two more relationships to consider:
l^2 - k^2 = 1
and
-2uv = 1

But, in any event, there are many (infinite) examples of where P & Q are members of R, but where PQ, and/or QP are NOT members of R.
Is R a CLOSED group, under multiplication?
Otherwise, your last equation is not necessarily true.

But the 2 relationships I've given ARE true.
Last edited by begbie68; 1 week ago
0
reply
DFranklin
Badges: 18
Rep:
?
#3
Report 1 week ago
#3
(Original post by begbie68)
Is R a CLOSED group, under multiplication?
Otherwise, your last equation is not necessarily true.

But the 2 relationships I've given ARE true.
R is closed under multiplication (it's the set of matrices of determinant 1). I think it's very likely the OP is trying to prove this (without refering to determinants).

Edit: except in that case, Q doesn't look right. OP needs to post the question, or at least explain what they're actually trying to do.
Last edited by DFranklin; 1 week ago
0
reply
bigmansouf
Badges: 20
Rep:
?
#4
Report Thread starter 1 week ago
#4
I am sorry for the confusion I thought that I wrote the question correctly the first time. I think it due to having posting issues on TSR
the full question is

Matrices P and Q are members of a set R which is defined as follows:
 R= \left \{ \begin{pmatrix} a &b \\  c &d  \end{pmatrix}: a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}
Prove that the product PQ is also a member of set R.
R is closed under multiplication (it's the set of matrices of determinant 1). I think it's very likely the OP is trying to prove this (without refering to determinants).

Edit: except in that case, Q doesn't look right. OP needs to post the question, or at least explain what they're actually trying to do.
0
reply
ghostwalker
  • Study Helper
Badges: 16
#5
Report 1 week ago
#5
(Original post by bigmansouf)
Question:
Matrices P and Q are members of a set R which is defined as follows:
 R= \left \{ \begin{pmatrix} a &b \\  c &d  \end{pmatrix}: a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}
Prove that the product PQ is also a member of set R.

Solution
let
 P=\begin{pmatrix} l &k \\  k & l \end{pmatrix}
 Q = \begin{pmatrix} u & v \\ u & -v \end{pmatrix}

QP=\begin{pmatrix} (lu+kv) & (lv-ku) \\ (ku+lv)&(kv-lu)\end{pmatrix}
since  ad-bc=1
({kv+lu)(kv-lu)}{(ku+lv)(lv-ku)=1}
(kv^2-lu^2)-(lv^2-ku^2)=1
(v^2+u^2)(k-l)=1
(v^2+u^2)=k-l
this is the part i am stuck at
please help me
Your problem is primarily with the general argument.

P,Q should be general/arbitrary elements of R. As it stands you've given them additional properties, so they're based on 2 parameters each, rather than 4.

Then you've looked at the product and said, "since ad-bc=1". Well this is what you're trying to prove, so you can't assume it.

You need to evaluate the ad-bc (when applied to the product PQ) and show that it is equal to 1. All you have to go on is that that equation holds true for the original matrices, since they are elements or R.

On a minor note, you've looked at QP, rather than PQ - don't know if that's just a typo.
Last edited by ghostwalker; 1 week ago
reply
DFranklin
Badges: 18
Rep:
?
#6
Report 1 week ago
#6
In addition to what ghostwalker has posted, for your matrix R, ad-bc is called the determinant of R (written det R) and there's a general result that det PQ = det P det Q (*)

Your problem boils down to showing '' if det P = det Q =1, then det PQ =1". But I suspect it's actually easier to prove (*), even though it's a more general result. It will make it a lot more obvious where you're trying to go I think.
0
reply
bigmansouf
Badges: 20
Rep:
?
#7
Report Thread starter 5 days ago
#7
(Original post by ghostwalker)
Your problem is primarily with the general argument.

P,Q should be general/arbitrary elements of R. As it stands you've given them additional properties, so they're based on 2 parameters each, rather than 4.

Then you've looked at the product and said, "since ad-bc=1". Well this is what you're trying to prove, so you can't assume it.

You need to evaluate the ad-bc (when applied to the product PQ) and show that it is equal to 1. All you have to go on is that that equation holds true for the original matrices, since they are elements or R.

On a minor note, you've looked at QP, rather than PQ - don't know if that's just a typo.
(Original post by DFranklin)
In addition to what ghostwalker has posted, for your matrix R, ad-bc is called the determinant of R (written det R) and there's a general result that det PQ = det P det Q (*)

Your problem boils down to showing '' if det P = det Q =1, then det PQ =1". But I suspect it's actually easier to prove (*), even though it's a more general result. It will make it a lot more obvious where you're trying to go I think.
let

 P=\begin{pmatrix} l &k \\ k & l \end{pmatrix}

 Q = \begin{pmatrix} z& y \\ x & w \end{pmatrix}
since  P=\begin{pmatrix} l &k \\ m & n \end{pmatrix}

thus  kn -lm =1
since Q = \begin{pmatrix}  z & y\\ x & w \end{pmatrix}

thus  zw - xy = 1
PQ=\begin{pmatrix} (kz+lx) & (ky+lw) \\ (mz+nx)&(my+nw)\end{pmatrix}

 = (kn-lm)(zw-xy)

but  kn -lm =1 and  zw - xy = 1

therefore;

 = (kn-lm)(zw-xy) = 1 \times 1 = 1



therefore. PQ is a member of set R
0
reply
DFranklin
Badges: 18
Rep:
?
#8
Report 5 days ago
#8
(Original post by bigmansouf)
let

 P=\begin{pmatrix} l &k \\ k & l \end{pmatrix}

 Q = \begin{pmatrix} z& y \\ x & w \end{pmatrix}
since  P=\begin{pmatrix} l &k \\ m & n \end{pmatrix}

thus  kn -lm =1
since Q = \begin{pmatrix}  z & y\\ x & w \end{pmatrix}

thus  zw - xy = 1
It looks like you have a bunch of typoes up to here; to some extent I'll give you the benefit of the doubt that it's TeX errors, but a lot of the above doesn't make sense.

PQ=\begin{pmatrix} (kz+lx) & (ky+lw) \\ (mz+nx)&(my+nw)\end{pmatrix}

 = (kn-lm)(zw-xy)
You started with a matrix, and you've now said it equals a number. This makes no sense!

Even if you meant that "the 'ad-bc' value for this matrix = (kn-lm)(zw-xy)", you've done absolutely nothing to indicate why this is true. This is the only part of the question that I think you can actually "earn marks for" (everything else is just restating definitions), so you absolutely need to actually have some workings here.

On a slightly different note: why on earth have you chosen that order for the letters in the matrices?

Surely P = \begin{pmatrix}k & l\\m & n\end{pmatrix}

and Q = \begin{pmatrix} x & y \\z & w\end{pmatrix} (or Q = \begin{pmatrix} w & x \\y & z\end{pmatrix}) would make more sense?
0
reply
ghostwalker
  • Study Helper
Badges: 16
#9
Report 5 days ago
#9
(Original post by bigmansouf)
...
R
It's pretty clear you haven't proof read that, with so many elements the wrong way round or just wrong, it's impossible to judge. And to jump from PQ to writing the product of two components needs some lines inbetween.
Last edited by ghostwalker; 5 days ago
reply
bigmansouf
Badges: 20
Rep:
?
#10
Report Thread starter 5 days ago
#10
(Original post by DFranklin)
It looks like you have a bunch of typoes up to here; to some extent I'll give you the benefit of the doubt that it's TeX errors, but a lot of the above doesn't make sense.

You started with a matrix, and you've now said it equals a number. This makes no sense!

Even if you meant that "the 'ad-bc' value for this matrix = (kn-lm)(zw-xy)", you've done absolutely nothing to indicate why this is true. This is the only part of the question that I think you can actually "earn marks for" (everything else is just restating definitions), so you absolutely need to actually have some workings here.

On a slightly different note: why on earth have you chosen that order for the letters in the matrices?

Surely P = \begin{pmatrix}k & l\\m & n\end{pmatrix}

and Q = \begin{pmatrix} x & y \\z & w\end{pmatrix} (or Q = \begin{pmatrix} w & x \\y & z\end{pmatrix}) would make more sense?
(Original post by ghostwalker)
It's pretty clear you haven't proof read that, with so many elements the wrong way round or just wrong, it's impossible to judge. And to jump from PQ to writing the product of two components needs some lines inbetween.
let
Since P and Q are members of a set R which is defined as follows;
 R = \left \{ \begin{pmatrix} a & b\\  c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc =1 \right \}
let
P=\left \{  \begin{pmatrix} k &l \\ m & n \end{pmatrix} k,l,m,n \epsilon \mathbb{R}, kn-lm = 1 \right \}

Q= \left \{ \begin{pmatrix} x&y \\ z & w \end{pmatrix} z,y,x,w \epsilon \mathbb{R}, xw-zy = 1 \right \}

Both det(Q) and det(P) is 1.
The product PQ=\begin{pmatrix} (kx+lz) & (ky+lw) \\ (mx+nz)&(my+nw)\end{pmatrix}

det(PQ)  = (kn-lm)(xw-zy)

but det(P) is  kn -lm =1 and det(Q) is  xw - zy = 1

therefore;

det(PQ)  = (kn-lm)(xw-zy) = 1 \times 1 = 1
looking at  R = \left \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc =1 \right \}

 \left \{ PQ= \begin{pmatrix} (kx+lz) &(ky+lw) \\ (mx+nz)&(my+nw)\end{pmatrix} k,l,m,n, z,y,x,w \epsilon \mathbb{R}, (kn-lm)-(xw-zy)= ad-bc =1 \right \}
therefore PQ is a member of set R where set R =  \left \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc = (kn-lm)(xw-zy) = 1 \right \}


I have tried my best i have proof read it. please point out where i went wrong. Writing a maths proof solution is my weakness - i am trying to improve it
Last edited by bigmansouf; 5 days ago
0
reply
begbie68
Badges: 16
Rep:
?
#11
Report 5 days ago
#11
what does the line beginning "determinant of PQ ..." mean? what do you ACTUALLY need to do to find det(PQ)??

If you think that writing a maths proof mostly involves getting your layout looking nice and neat and a flourish of "therefore ...<copy out the thing you had been given to prove> ..." then you need to change your focus a little. These things CAN matter (but only very very slightly).

More importantly, you MUST get your algebra right.
0
reply
ghostwalker
  • Study Helper
Badges: 16
#12
Report 5 days ago
#12
(Original post by bigmansouf)
let
Since P and Q are members of a set R which is defined as follows;
 R = \left \{ \begin{pmatrix} a & b\\  c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc =1 \right \}
let
P=\left \{  \begin{pmatrix} k &l \\ m & n \end{pmatrix} k,l,m,n \epsilon \mathbb{R}, kn-lm = 1 \right \}

Q= \left \{ \begin{pmatrix} x&y \\ z & w \end{pmatrix} z,y,x,w \epsilon \mathbb{R}, xw-zy = 1 \right \}

Both det(Q) and det(P) is 1.
If you're going to use determinants and their properties, then it's a two line proof:

Let P,Q be elements of R, then det(P)=det(Q) = 1.
Since det(PQ)=det(P)det(Q), it follows det(PQ)=1 and so PQ is an element of R.

HOWEVER, it's clear from the question, that you're expected not to use determinants or their properties. They're expecting you to work the algebra from the definition of the set.

You need so show, that when the formula ad-bc is applied to the matrix PQ, then the result is 1 - as DFranklin said in his previous post.
reply
bigmansouf
Badges: 20
Rep:
?
#13
Report Thread starter 5 days ago
#13
(Original post by ghostwalker)
If you're going to use determinants and their properties, then it's a two line proof:

Let P,Q be elements of R, then det(P)=det(Q) = 1.
Since det(PQ)=det(P)det(Q), it follows det(PQ)=1 and so PQ is an element of R.

HOWEVER, it's clear from the question, that you're expected not to use determinants or their properties. They're expecting you to work the algebra from the definition of the set.

You need so show, that when the formula ad-bc is applied to the matrix PQ, then the result is 1 - as DFranklin said in his previous post.
let
P= \begin{pmatrix} k &l \\ m & n \end{pmatrix}
and
Q= \begin{pmatrix} x&y \\ z & w \end{pmatrix} be members of set R.

since P and Q are elements then  k=a, l=b, m=c , n=d
also,  x=a, y=b, z=c, w=d

By substitution;
 kn-lm =ad-bc = 1 and   xw-zy =1

 PQ = \begin{pmatrix} (kx+lz) &(ky+lw) \\ (mx+nz)&(my+nw)\end{pmatrix}

 (kn-lm)(xw-zy)=  (ad-bc)(ad-bc)=1

Since  (kn-lm)(xw-zy) =1 , PQ is a member of R.
is this the correct way?
0
reply
DFranklin
Badges: 18
Rep:
?
#14
Report 5 days ago
#14
(Original post by bigmansouf)
 PQ = \begin{pmatrix} (kx+lz) &(ky+lw) \\ (mx+nz)&(my+nw)\end{pmatrix}

 (kn-lm)(xw-zy)=  (ad-bc)(ad-bc)=1
What is this last equation even supposed to mean? It has no context, no explanation.

Since  (kn-lm)(xw-zy) =1 , PQ is a member of R.
is this the correct way?
No.

You need to show that "ad-bc" for PQ equals 1 (or more generally, that it equals (kn-lm)(xw-zy)).

That is: you need to evaluate (kx+lz)(my+nw) - (ky+lw) (mx+nz) and show it equals 1 (or equals (kn-lm)(xw-zy)).
Last edited by DFranklin; 5 days ago
0
reply
bigmansouf
Badges: 20
Rep:
?
#15
Report Thread starter 4 days ago
#15
(Original post by DFranklin)
What is this last equation even supposed to mean? It has no context, no explanation.

No.

You need to show that "ad-bc" for PQ equals 1 (or more generally, that it equals (kn-lm)(xw-zy)).

That is: you need to evaluate (kx+lz)(my+nw) - (ky+lw) (mx+nz) and show it equals 1 (or equals (kn-lm)(xw-zy)).
So the crucial step i am missing is
 (kx+lz)(my+nw) - (ky+lw) (mx+nz) = (kxmy+kxnw+lzmy+lznw) -(kymx+kynz+lwmx+lwnx)
= kxmy-kymx+kxnw-kynz+lzmy-lwmx+ lznw-lwnz=  kxnw-kynz+lzmy-lwmx= kn(xw-zy)+lm(zy-xw)
=kn(xw-zy)-lm(xw-zy)=(kn-lm)(xw-zy)

since (kn-lm) = ad-bc and xw-z)y = ad-bc
so  (kn-lm)(xw-zy) = (ad-bc)(ad-bc)= 1
since  (kx+lz)(my+nw) - (ky+lw) (mx+nz) = 1 , PQ is a member of R

in terms of the evaluating  (kx+lz)(my+nw) - (ky+lw) (mx+nz) i did evaluate it on pen and paper but i did not write it on tsr because i thought it was not a necessary step I should include
I am sorry if this is what you have been trying to tel me the whole time
Last edited by bigmansouf; 4 days ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
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

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (246)
39.42%
No - but I will (42)
6.73%
No - I don't want to (45)
7.21%
No - I can't vote (<18, not in UK, etc) (291)
46.63%

Watched Threads

View All
Latest
My Feed