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Thread: matrices problem

  1. #1
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    matrices problem

    Question:
    A transformation T is represented by
    $\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $ where $\displaystyle b \epsilon \mathbb{R}^{+} $
    a) Draw a diagram showing the unit square and its image under T
    [b) Show that area is invariant under T
    c) Show that T maps the curve $\displaystyle y =1/x $ onto itself
    d) Show that T maps the region bounded by the curve $\displaystyle y = 1/x $, the lines x=1 and x =a and the x-axis onto the region bounded by the same curve the lines x =b and x =ab and the x-axis.
    e) hence show that $\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx $
    f)Given that $\displaystyle F(t) = \int_{1}^{t} \frac{1}{x} dx $ show that $\displaystyle F(ab)=F(a)+F(b) $

    my attempt:a)
    the unit square $\displaystyle = \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix} $
    transformed by T ;

    $\displaystyle \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} b &0 \\ 0 & 1/b \end{pmatrix} \begin{pmatrix} 0 &1 & 1 &0 \\ 0& 0 & 1 &1 \end{pmatrix} = \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix} 0 &b &b & 0\\ 0 &0 & 1/b &1/b \end{pmatrix} $

    matrices problem-m1.jpg

    did i draw the right diagram?

    b) the area is invariant because the area of the transformed image is $\displaystyle \frac{1}{b} \times b = 1 $

    the determinant of the unit square = 1
    Since the area of the transformed = 1 and the determinant of the unit square = 1 therefore, the area is invariant.

    c)$\displaystyle \begin{pmatrix}x'\\ y'\end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $

    since it maps $\displaystyle y = \frac{1}{x}$ onto itself thus

    $\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}b &0 \\ 0 & 1/b \end{pmatrix}\begin{pmatrix}x \\ \frac{1}{x}\end{pmatrix} $

    $\displaystyle \begin{pmatrix}x' \\ \frac{1}{x}' \end{pmatrix} = \begin{pmatrix}bx\\ \frac{b}{x}\end{pmatrix} $


    T maps points on y =1/x to new points on the curve y =1/x

    d) I need help this part i dont know what to do
    e)
    $\displaystyle \int_{1}^{a} \frac{1}{x} dx = \int_{a}^{ab} \frac{1}{x} dx $
    we know that $\displaystyle \int \frac{1}{x}dx = ln \left | x \right |+c $
    thus, $\displaystyle \int_{1}^{a} \frac{1}{x} dx = ln \left | a \right | - ln \left | 1 \right | =ln \left | a \right | - 0 = ln \left | a \right |$
    $\displaystyle \int_{a}^{ab} \frac{1}{x} dx = ln \left | ab \right | - ln \left | a \right | = \frac{ln \left | ab \right |}{ln \left | a \right |}= \frac{ln \left | a \right |ln \left | b \right |}{ln \left | a \right |}= ln \left | b \right | $

    a =b
    i am not confident if i did this part right. I would like to know if i did this part correctly
    f) can i please have some help doing part F?
    Please I would like help with this question
    Last edited by bigmansouf; May 9th 2019 at 09:44 PM. Reason: to ask stuff
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  2. #2
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    Re: matrices problem

    In your second last line in (e) you are saying that ln|ab| - ln |a| = ln|ab|/ln|a|, in other words the subtraction of logs is the same as division of logs?? Rethink that!


    The rule is: ln(m/n) = ln m - ln n . Can you see what you have done wrong? It is a common mistake!
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