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Math Help - Matrix, find variable

  1. #1
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    Matrix, find variable

    For some reason im just not getting this, if anyone could help me solve this problem, explaining exactly how they did it, i would realy appreciate it.

    7x + 6y - 3z = -7
    7x - 7y + kz = 4
    7x - 20y - 3z = 10

    For the system to be consistent k must not = ?
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  2. #2
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    Quote Originally Posted by Juggalomike View Post
    For some reason im just not getting this, if anyone could help me solve this problem, explaining exactly how they did it, i would realy appreciate it.

    7x + 6y - 3z = -7
    7x - 7y + kz = 4
    7x - 20y - 3z = 10

    For the system to be consistent k must not = ?
    Write the system in matrix form...

    \left[\begin{matrix} 7 & 6 & -3 \\ 7 & -7 & k \\ 7 & -20 & -3\end{matrix}\right] \left[\begin{matrix} x \\ y \\ z \end{matrix}\right] = \left[\begin{matrix} -7 \\ 4 \\ 10 \end{matrix}\right].

    I'm going to rewrite this equation as

    A\mathbf{x} = \mathbf{b}.

    Using some matrix algebra

    A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}

    I\mathbf{x} = A^{-1}\mathbf{b}

    \mathbf{x} = A^{-1}\mathbf{b}.


    So does it make sense that to have a solution, you need to be able to premultiply both sides by the inverse matrix of A. Therefore this inverse matrix has to exist.

    A^{-1} exists when |A| \neq 0.

    So evaluate |A|, and solve for k when |A| = 0

    This will tell you the value(s) of k that you can not have.
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  3. #3
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    I tried solving the sum ..Can i know if k should not be = -3 is correct..
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  4. #4
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    Let's see.

    For a 3 \times 3 matrix A = \left[\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right]

    then |A| = aei + bfg + cdh - ceg - afh - bdi.


    So in your case

    |A| = 7(-7)(-3) + 6(k)(7) + (-3)(7)(-20) - (-3)(-7)(7) - 7(k)(-20) - 6(7)(-3)

     = 147 + 42k + 420 - 147 + 140k + 126

     = 546 + 182k


    Solving |A| = 0 for k gives

    546 + 182k = 0

    182k = -546

    k = -3.


    So yes, you can not let k = -3
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