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Math Help - Solve System of Equations with A^-1

  1. #1
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    Solve System of Equations with A^-1

    Write the system in the vector matrix form Ax=b and solve by finding A^-1

    Vector Matrix form?
    [2,1;4,3]*[x;y] = [2,-4]

    2x+y=2
    4x+3x=-4

    So A^-1 =
    [1.5,-.5;-2,1]

    So I don't understand what I am suppose to do with the A^-1. I understand how to use ref, rref and Cramer's rule to solve these equations but I am not understanding what method they are wanting me to use.
    Reference: Answer is x=5, y=-8
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  2. #2
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    Re: Solve System of Equations with A^-1

    Quote Originally Posted by bustacap09 View Post
    Write the system in the vector matrix form Ax=b and solve by finding A^-1

    Vector Matrix form?
    [2,1;4,3]*[x;y] = [2,-4]

    2x+y=2
    4x+3x=-4

    So A^-1 =
    [1.5,-.5;-2,1]

    So I don't understand what I am suppose to do with the A^-1. I understand how to use ref, rref and Cramer's rule to solve these equations but I am not understanding what method they are wanting me to use.
    Reference: Answer is x=5, y=-8
    Well you have Au = v. You found A^{-1}, so then A^{-1}Au = A^{-1}v implies u = A^{-1}v.

    Does that answer the question or am I not getting where you are confused?

    -Dan
    Thanks from bustacap09
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  3. #3
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    Re: Solve System of Equations with A^-1

    "Well you have Au = v. You found A^{-1}, so then A^{-1}Au = A^{-1}v implies u = A^{-1}v."

    I didn't understand how you used the A^-1 to find the values for [x,y]. But this makes perfect sense.

    Thank you very much.
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  4. #4
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    Re: Solve System of Equations with A^-1

    If A\overline{x}=\overline{b}, then \overline{x}=A^{-1}\overline{b}. So, \left(\begin{matrix}{\frac{3}{2} & -\frac{1}{2}\\-2 & 1\end{matrix}\right)\left(\begin{matrix}2\\-4\end{matrix}\right)=\left(\begin{matrix}5\\-8\end{matrix}\right). Straight up plug and chuggin'.
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