Matrix problem

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.

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.

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 $zw - xy = 1$

...
R

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?

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.

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.

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

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

Since $(kn-lm)(xw-zy) =1$, PQ is a member of R.
is this the correct way?

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)).