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Math Help - matrix-- determinants

  1. #1
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    matrix-- determinants

    hello anyone, i have a problem on the determinant
    here are 3 liner equations:
    y=3
    y-x=0
    y-3x-6=0
    there has a point (3,3) which can be passed through by these 3 lines
    the determinant of the coefficients of these three lines is zero.
    why? Is it always true if any 3 lines are concurrent? Thank you!
    p.s please forgive my poor English.
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  2. #2
    A Plied Mathematician
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    You have an overdetermined, inconsistent system of equations there. There is no solution. If you had a plus 6 in the last equation instead of a minus 6, you'd have an overdetermined consistent system. I don't understand what you mean when you say 'The determinant of the coefficients of these three lines is zero". The determinant is only defined for square matrices, but you have an overdetermined system. Therefore, the determinant is not defined. Exactly what matrix are you taking the determinant of?
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  3. #3
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    The matrix of coefficients is \begin{bmatrix}0 & 3 & 3\\ -1 & 1 & 0 \\ -3 & 1 & 6\end{bmatrix}
    which, as Ackbeet says, does not have a determinant because it is not square.

    I thought for a moment you meant the augmented matrix, \begin{bmatrix}0 & 3 & 3\\ -1 & 1 & 0 \\ -3 & 1 & 6\end{bmatrix} which has a determinant, but its determinant is 24 not 0.


    What you are asking about is simply not true.
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  4. #4
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    I am sorry. I think I got some mistakes on my post. I will write it again on below, please read it.
    here are three equation:
    t^(3)x-ty+3-3t^(4)=0
    tx-t^(3)y +3-3t^(4)=0
    x+y=0
    my teacher consider =0
    so he said there three equation are concurrent
    but i don't understand.
    Last edited by kenpoon; December 10th 2010 at 07:01 PM.
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  5. #5
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    Should the second equation have t^3y
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  6. #6
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    Assume you meant t^3y

    \displaystyle<br />
\begin{bmatrix}<br />
t^3 & -t & 3-3t^4\\ <br />
t & -t^3 & 3-3t^4\\ <br />
1 & 1 & 0<br />
\end{bmatrix}\rightarrow \ \mbox{rref}=\begin{bmatrix}<br />
1 & 0 & \frac{3(t^2-1)}{t}\\ <br />
0 & 1 & \frac{3(t^2-1)}{t}\\ <br />
0 & 0 & 0<br />
\end{bmatrix}\rightarrow \ x=-\frac{3(t^2-1)}{t}, \ y=-\frac{3(t^2-1)}{t}
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  7. #7
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    yes
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