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Math Help - Matrices

  1. #1
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    Matrices

    <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 0 & 14\;\\\;0 & 0 & 1 & -13\end{vmatrix}<br />

    If the variables in the equations of this matrix were x, y, and z:

    x = 14
    z = -13

    I'm just not sure what to answer for y. Does having no y value in the augmented matrix mean the system has infinitely many solutions or a unique solution?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Thomas View Post
    <br />
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 0 & 14\;\\\;0 & 0 & 1 & -13\end{vmatrix}<br />

    If the variables in the equations of this matrix were x, y, and z:

    x = 14
    z = -13

    I'm just not sure what to answer for y. Does having no y value in the augmented matrix mean the system has infinitely many solutions or a unique solution?
    there is no leading 1 in the y column, it means we can equate y to a parameter (thus there are infinitely many solutions, because the parameter can be essentially anything, and each new parameter gives rise to a new solution set)... you know this
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  3. #3
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    Thank you for the help.

    For some reason, when doing questions that actually mean something, I start doubting my instincts.
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