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Math Help - Matrices

  1. #1
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    Matrices

    A transformation in the plane is represented by a 2 x 2 matrix.
    Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2)

    Can someone help me to find the transformation matrix for this?
    I dont have any working to show as i dont know where to start

    Thanks
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  2. #2
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    Re: Matrices

    Hello, MMCS!

    A transformation in the plane is represented by a 2 x 2 matrix.
    Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2).
    Find the transformation matrix.

    You can start by letting the transformation matrix be: . A \;=\;\begin{bmatrix} a&b \\ c&d\end{bmatrix}

    Then: . (7,7)\begin{bmatrix}a&b\\c&d\end{bmatrix} \;=\;(\text{-}5,\text{-}6) \quad\Rightarrow\quad \begin{Bmatrix}7a + 7c &=& \text{-}5 \\ 7b + 7d &=& \text{-}6 \end{Bmatrix}

    Also: . (\text{-}2,8)\begin{bmatrix}a&b\\c&d\end{bmatrix}\;=\;( \text{-}4,2) \quad\Rightarrow\quad \begin{Bmatrix}\text{-}2a + 8c &=& \text{-}4 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}


    We have two systems of equations:

    . . . . . . . . . . . \begin{Bmatrix}7a + 7c &=& \text{-}5 \\ \text{-}2a+8c &=& \text{-}4\end{Bmatrix} \qquad\begin{Bmatrix}7b+7d &=& \text{-}6 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}

    Their solutions are: . \begin{Bmatrix} a&=& \text{-}\frac{6}{35} \\ \\[-4mm]   c &=& \text{-}\frac{19}{35} \end{Bmatrix} \qquad \begin{Bmatrix}b &=& \text{-}\frac{31}{35} \\ \\[-4mm]    d &=& \frac{11}{35} \end{Bmatrix}


    Therefore: . A \;=\;\begin{bmatrix}\text{-}\frac{6}{35} & \text{-}\frac{31}{35} \\ \\[-3mm] \text{-}\frac{19}{35} & \frac{1}{35} \end{bmatrix}
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  3. #3
    Super Member ILikeSerena's Avatar
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    Re: Matrices

    Here's an alternative method:

    We have:

    A \begin{pmatrix}7 \\ 7\end{pmatrix} = \begin{pmatrix}-5 \\ -6\end{pmatrix} \qquad A \begin{pmatrix}-2 \\ 8\end{pmatrix} = \begin{pmatrix}-4 \\ 2\end{pmatrix}

    Therefore:

    A \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix} = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix}

    A = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix} \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix}^{-1}
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  4. #4
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    Re: Matrices

    Yet a third way: (1, 0)= a(7, 7)+ b(-2, 8) for some a and b. That gives the two equations 7a- 2b= 1 and 7a+ 8b= 0. If we subtract the first equation from the second we get 10b= -1 so that b= -1/10. Putting that back into the first equation, 7a+ 2/10= 1 so 7a= 8/10 and a= 8/70. That is (1, 0)= (8/70)(7, 7)+ (-1/10)(-2, 8) and then A(1, 0)= (8/70)A(7, 7)+ (-1/10)A(-2, 8)= (8/70)(-5, -6)+ (-1/10)(-4, 2)= (8/70, -31/35). The point of reducing to (1, 0) is that \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}. That is the first column of the matrix is whatever A maps (1, 0) to. This tells us that the first column of the matrix is \begin{bmatrix}\frac{8}{70} \\ -\frac{31}{35}\end{bmatrix}.

    Now, find a, b such that a(7, 7)+ b(-2, 8)= (0, 1) to find the second column.
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