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Math Help - Matrices - Linear equations

  1. #1
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    Matrices - Linear equations



    I don't quite understand how to use that augmented matrix to prove anything about a, b and c.
    Any ideas on where to start?
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  2. #2
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    The above image seems a bit hard to read.
    Here is another question similar in format, though, and it's clearer:


    It's part of our matrices unit but I'm not sure how to incorporate that into this question.
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  3. #3
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    25. You know four points that lie on the curve, \displaystyle (0, 10), (1, 7), (3, -11), (4, -14).

    So that means they all satisfy the equation \displaystyle y = a\,x^3 + b\,x^2 + c\,x + d.


    So
    \displaystyle 10 = a\cdot 0^3 + b\cdot 0^2 + c\cdot 0 + d

    \displaystyle 7 = a\cdot 1^3 + b\cdot 1^2 + c\cdot 1 + d

    \displaystyle -11 = a\cdot 3^3 + b\cdot 3^2 + c\cdot 3 + d

    \displaystyle -14 = a\cdot 4^3 + b\cdot 4^2 + c\cdot 4 + d.


    So now you can set up your matrix equation...

    \displaystyle \left[\begin{matrix}0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1\\ 27 & 9 & 3 & 1 \\ 64 & 16 & 4 & 1\end{matrix}\right]\left[\begin{matrix}a \\ b\\ c\\d \end{matrix}\right] = \left[\begin{matrix}\phantom{-}10 \\ \phantom{-}7 \\ -11 \\ -14\end{matrix}\right]
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    .I usually write the constants in the same brackets as the large matrix here (to the furthest right side of the matrix, since that's what we were taught). So:

    0 0 0 1 10
    1 1 1 1 7
    27 9 3 1 -11
    64 16 4 1 -14
    Is that right?

    Would I use Gauss-Jordan to reduce that into reduced row-echelon?
    I'm still confused about how to find the variables from there.
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    [QUOTE=TN17;604092].I usually write the constants in the same brackets as the large matrix here (to the furthest right side of the matrix, since that's what we were taught). So:

    0 0 0 1 10
    1 1 1 1 7
    27 9 3 1 -11
    64 16 4 1 -14
    Is that right?/[quote]
    Yes, that is what is meant by the "augmented" matrix.

    Would I use Gauss-Jordan to reduce that into reduced row-echelon?
    I'm still confused about how to find the variables from there.
    Yes, you certainly could use Gauss-Jordan. I would be inclined to swap the first row to the last initially to get
    \begin{bmatrix}1 & 1 & 1 & 1 & 7 \\ 27 & 9 & 3 & 1 & -11\\ 64 & 16 & 4 & 1 & -14 \\ 0 & 0 & 0 & 1 & -14\end{bmatrix}
    Then subtract 27 times the first row from the second, 64 times the first row from the third, etc.
    Last edited by HallsofIvy; January 13th 2011 at 09:20 AM.
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    If you have a CAS, it's quite simple if you know that the solution to the matrix equation

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

    is

    \displaystyle \mathbf{x} = \mathbf{A}^{-1}\mathbf{b}...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    If you have a CAS, it's quite simple if you know that the solution to the matrix equation

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

    is

    \displaystyle \mathbf{x} = \mathbf{A}^{-1}\mathbf{b}...
    On the other hand, if you want to actually learn some linear algebra there is no better way than doing it by hand!
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    On the other hand, if you want to actually learn some linear algebra there is no better way than doing it by hand!
    I disagree, using matrix algebra to solve matrix equations and then programming a computer/CAS to perform the tedious calculations that you are almost certain to make a mistake with is far more beneficial.
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  10. #10
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    I'll meet you guys halfway. To maximise understanding the OP should know how to solve a system like the one above by hand. Once confident with such methods then use technology to save time.
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