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Math Help - Inverse

  1. #1
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    Inverse

    Set up a system belw as a matrix multiplication of the form AB=C. Find the inverse A^-1 of A and use it to solve the system.

    3x + -6y + -17z = 288

    11x + -17y + 110z = 79

    -5x + 7y + 8z = 20.

    I am able to get the first part of the matrix but I get lost when it comes to working on the second row.
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  2. #2
    Moo
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    Hello,

    \begin{pmatrix} 3&-6&-17 \\ 11&-17&110 \\ -5&7&8 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 288 \\ 79 \\ 20 \end{pmatrix}

    This is your product.

    After that, I don't know which methods you've used to calculate the inverse of a 3x3 matrix.
    Here is the most common one :

    Start with :
    \left( \begin{array}{ccc|ccc} 3&-6&-17&1&0&0 \\ 11&-17&110&0&1&0 \\ -5&7&8&0&0&1 \end{array} \right)


    And make row operations to find the ?'s in this matrix :
    \left( \begin{array}{ccc|ccc} 1&0&0&?&?&? \\ 0&1&0&?&?&? \\ 0&0&1&?&?&? \end{array} \right)


    The ?'s will form A^{-1}



    After that, get back to the equation :
    AB=C

    multiply on the left by A^{-1} :

    \underbrace{A^{-1}A}_{=I}B=A^{-1}C

    B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}=A^{-1}C=A^{-1}\begin{pmatrix} 288 \\ 79 \\ 20 \end{pmatrix}
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