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Thread: matrices problem

  1. #1
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    matrices problem

    Question: The matrix M is given by M $\displaystyle = \begin{pmatrix} a & b\\ c & d \end{pmatrix} $, where $\displaystyle a,b,c,d \epsilon \mathbb{R} $.

    a)Find $\displaystyle M^{2} $.

    b)Given that $\displaystyle M^{2} = M $ and that b and c are non-zero, prove that M is singular.

    c)Prove that in this case, the transformation T, is defined by $\displaystyle T:\begin{pmatrix} x\\ y \end{pmatrix} \mapsto M \begin{pmatrix} x\\ y \end{pmatrix} $

    maps all points of the plane to points of the line $\displaystyle (1-a) x =by $



    my attempt:

    $\displaystyle M^2 = \begin{pmatrix} a^2 + bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix} $



    prove that M is singular

    $\displaystyle \begin{pmatrix} a^2 + bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

    if matrix is singular then $\displaystyle det(M) = 0 $



    thus we need to prove that $\displaystyle det(M) = 0 $

    $\displaystyle det(M) = ad -bc $



    $\displaystyle a^2+bc=a......1 \\ ab+bd = b .....2\\ ac+dc=c ......3 \\ bc+d^2=d .....4 $



    from 1

    $\displaystyle bc = a- a^2 $





    from 2

    $\displaystyle ab+ bd =b $



    $\displaystyle bd = b - ab $

    $\displaystyle bd = b(1-a) $

    $\displaystyle d = 1-a $

    sub bc and d into $\displaystyle ad-bc $



    thus

    $\displaystyle a(1-a) - (a - a^2) = 0 $



    since $\displaystyle ad -bc =0$ therefore M is singular



    I need help with part c

    please help
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  2. #2
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    Re: matrices problem

    the image of $T$ is a straight line thru $(0,0)$

    and

    $\displaystyle T(0,1)=(b,1-a)$
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  3. #3
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    Re: matrices problem

    "b)Given that M^2=M and that b and c are non-zero, prove that M is singular."
    From M^2= M, M^2- M= M(M-I)= 0. That is sufficient to show that M is singular.
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    Re: matrices problem

    Quote Originally Posted by Idea View Post
    the image of $T$ is a straight line thru $(0,0)$

    and

    $\displaystyle T(0,1)=(b,1-a)$
    sorry but could you explain how you found that the image of $T$ is a straight line thru $(0,0)$
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    Re: matrices problem

    Quote Originally Posted by bigmansouf View Post
    sorry but could you explain how you found that the image of $T$ is a straight line thru $(0,0)$
    ImT is a subspace of $R^2$ so it is either trivial, a line thru the origin ( since T(0,0)=(0,0) ),

    or the whole plane depending on rank T = 0,1, or 2

    T is singular since M is singular, so T is not surjective and $T \neq 0$ therefore

    ImT = straight line i.e. dim ImT = 1 and ImT is spanned by the vector $T(0,1)=(b,1-a)$

    in a coordinate system that is the line $(1-a)x=b y$
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