1. ## matrices problem

Question:
Matrices P and Q are members of a set R which is defined as follows:
$\displaystyle R= R= \left \{ \begin{pmatrix} a &b \\ c &d \end{pmatrix} a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}$

Solution

$\displaystyle P=\begin{pmatrix} l &k \\ k & l \end{pmatrix}$
$\displaystyle Q = \begin{pmatrix} u & v \\ u & -v \end{pmatrix}$

$\displaystyle QP=\begin{pmatrix} (lu+kv) & (lv-ku) \\ (ku+lv)&(kv-lu)\end{pmatrix}$
since $\displaystyle ad-bc=1$
$\displaystyle ({kv+lu)(kv-lu)}{(ku+lv)(lv-ku)=1}$
$\displaystyle (kv^2-lu^2)-(lv^2-ku^2)=1$
$\displaystyle (v^2+u^2)(k-l)=1$
$\displaystyle (v^2+u^2)=k-l$
this is the part i am stuck at

2. ## Re: matrices problem

I'm not seeing the question

3. ## Re: matrices problem

You state that P and Q are members of a certain set of matrices then give a "solution".

A solution to what problem? Is it to show that R is "closed under multiplication"? If you already have the theorem that "det(AB)= det(A)det(B)" then it is trivial!

4. ## Re: matrices problem

Originally Posted by HallsofIvy
You state that P and Q are members of a certain set of matrices then give a "solution".

A solution to what problem? Is it to show that R is "closed under multiplication"? If you already have the theorem that "det(AB)= det(A)det(B)" then it is trivial!
sorry about that the full question is
Question:
Matrices P and Q are members of a set R which is defined as follows:
$\displaystyle R= \left \{ \begin{pmatrix} a &b \\ c &d \end{pmatrix}: a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}$
Prove that the product PQ is also a member of set R.

5. ## Re: matrices problem

Originally Posted by bigmansouf
sorry about that the full question is
Question: Matrices P and Q are members of a set R which is defined as follows:
$\displaystyle R= \left \{ \begin{pmatrix} a &b \\ c &d \end{pmatrix}: a, b, c, d \epsilon \mathbb{R}, ad-bc=1\right \}$
Prove that the product PQ is also a member of set R.
I cannot see where your second post is any great improvement upon the first.

It appears that the set $\mathcal{R}$ is the set of $2\times 2$ matrices over the real field having the property that the determinate of each is one.
As Prof. Ivey has already pointed out, If $A~\&~B$ are square matrices of the same order then $|A\cdot B|=|A|\cdot |B|$
Think about it. If both have determinate one, what is the determinate of the product?

6. ## Re: matrices problem

Originally Posted by HallsofIvy
You state that P and Q are members of a certain set of matrices then give a "solution".

A solution to what problem? Is it to show that R is "closed under multiplication"? If you already have the theorem that "det(AB)= det(A)det(B)" then it is trivial!
I am sorry to be a bit of a pain but I have not yet come across the theorem you are speaking of. This question is under section A and the determinants of matrices in under section C in my textbook so I believe there is a way to do this without using that theorem which is what i am trying to find.

7. ## Re: matrices problem

I guess you are expected to brute force it

$P=\begin{pmatrix}p_a &p_b\\p_c &p_d\end{pmatrix}$
$Q=\begin{pmatrix}q_a &q_b\\q_c &q_d\end{pmatrix}$

$p_a p_d - p_c p_b = q_a q_d - q_c q_b = 1$

$PQ = \begin{pmatrix} p_a q_a+p_b q_c & p_a q_b+p_b q_d \\ p_c q_a+p_d q_c & p_c q_b+p_d q_d \\ \end{pmatrix}$

$|PQ| = -p_a q_b q_c p_d-q_a p_b p_c q_d+p_a q_a p_d q_d+p_b q_b p_c q_c =$

$\left(p_a p_d-p_b p_c\right) \left(q_a q_d-q_b q_c\right) = 1 \cdot 1 = 1$

8. ## Re: matrices problem

Originally Posted by romsek
I guess you are expected to brute force it

$P=\begin{pmatrix}p_a &p_b\\p_c &p_d\end{pmatrix}$
$Q=\begin{pmatrix}q_a &q_b\\q_c &q_d\end{pmatrix}$

$p_a p_d - p_c p_b = q_a q_d - q_c q_b = 1$

$PQ = \begin{pmatrix} p_a q_a+p_b q_c & p_a q_b+p_b q_d \\ p_c q_a+p_d q_c & p_c q_b+p_d q_d \\ \end{pmatrix}$

$|PQ| = -p_a q_b q_c p_d-q_a p_b p_c q_d+p_a q_a p_d q_d+p_b q_b p_c q_c =$

$\left(p_a p_d-p_b p_c\right) \left(q_a q_d-q_b q_c\right) = 1 \cdot 1 = 1$
thank you very much for helping me

i am trying to write the answer in a more proof format - i need to improve me proof writing solution so i can get full marks in exam
Please can you tell me if what i have written below is good

Thank you

Since P and Q are members of a set R which is defined as follows;

$\displaystyle R = \left \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc =1 \right \}$

let

$\displaystyle \left \{ P=\begin{pmatrix} k &l \\ m & n \end{pmatrix} k,l,m,n \epsilon \mathbb{R}, kn-lm = 1 \right \}$

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

The determinants of both Q and P is 1.

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

determinant of PQ $\displaystyle = (kn-lm)(xw-zy)$

but determinant of P is $\displaystyle kn -lm =1$ and determinant of Q is$\displaystyle xw - zy = 1$

therefore;

determinant $\displaystyle = (kn-lm)(xw-zy) = 1 \times 1 = 1$

looking at $\displaystyle R = \left \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc =1 \right \}$

$\displaystyle \left \{ PQ= \begin{pmatrix} (kx+lz) &(ky+lw) \\ (mx+nz)&(my+nw)\end{pmatrix} k,l,m,n, z,y,x,w \epsilon \mathbb{R}, (kn-lm)-(xw-zy)= ad-bc =1 \right \}$

therefore PQ is a member of set R where set R = $\displaystyle \left \{ \begin{pmatrix} a & b\\ c & d \end{pmatrix}: a,b,c,d \epsilon \mathbb{R}, ad-bc = (kn-lm)(xw-zy) = 1 \right \}$