Results 1 to 8 of 8
Like Tree8Thanks
  • 2 Post By romsek
  • 2 Post By HallsofIvy
  • 1 Post By Plato
  • 3 Post By romsek

Thread: matrices problem

  1. #1
    Member
    Joined
    May 2015
    From
    United Kingdom
    Posts
    114

    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
    please help me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,501
    Thanks
    2846

    Re: matrices problem

    I'm not seeing the question
    Thanks from topsquark and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,239
    Thanks
    3355

    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!
    Last edited by HallsofIvy; Apr 10th 2019 at 07:52 AM.
    Thanks from topsquark and bigmansouf
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2015
    From
    United Kingdom
    Posts
    114

    Re: matrices problem

    Quote Originally Posted by HallsofIvy View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,348
    Thanks
    3243
    Awards
    1

    Re: matrices problem

    Quote Originally Posted by bigmansouf View Post
    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?
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2015
    From
    United Kingdom
    Posts
    114

    Re: matrices problem

    Quote Originally Posted by HallsofIvy View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,501
    Thanks
    2846

    Re: matrices problem

    I guess you are expected to brute force it

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

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

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

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

    $\left(p_a p_d-p_b p_c\right) \left(q_a q_d-q_b q_c\right) = 1 \cdot 1 = 1$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2015
    From
    United Kingdom
    Posts
    114

    Re: matrices problem

    Quote Originally Posted by romsek View Post
    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 \} $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Matrices Problem?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 10th 2011, 03:08 PM
  2. Matrices problem (gps)?
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Jan 29th 2011, 03:10 AM
  3. Matrices problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Dec 7th 2010, 04:01 AM
  4. Matrices problem-
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Apr 29th 2009, 06:35 AM
  5. matrices problem...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 23rd 2009, 07:32 PM

/mathhelpforum @mathhelpforum