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Math Help - Matrices

  1. #1
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    Matrices

    Matrix A = 8 -1
    4 -3

    Matrix B= x y z
    u v w

    Answer = 49 4 24
    27 12 12

    I need to get the answers for x,y,z,u,v and w by using Matrix A and the already given answer, anyone understand it?
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  2. #2
    Newbie Esteban's Avatar
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    Re: Matrices

    that's all? I really don't know what the exercise is trying to mean XD
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  3. #3
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    Re: Matrices

    Yeah all it says is find x,y,z,u,v and w. I was thinking taking the inverse of A and using it but I'm not quite sure how or if that would work.
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  4. #4
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    Re: Matrices

    Unless you are told that there is some relationship between A and B, that's impossible.

    That's like saying "y= 4. Find x"!!
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  5. #5
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    Re: Matrices

    Ok, thank you!
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  6. #6
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    Re: Matrices

    Actually I figured it out, if you take the inverse of A and multiply it by the finished matrix you get the answer.
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  7. #7
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    Re: Matrices

    Hello, JP2012!

    You forgot to give us the set-up to the problem.


    \begin{bmatrix} 8&\text{-}1 \\ 4&\text{-}3\end{bmatrix} \cdot \begin{bmatrix} x&y&z \\ u&v&w \end{bmatrix}\;=\; \begin{bmatrix}49&4&24 \\ 27&12&12\end{bmatrix}

    \text{Find matrix }\begin{bmatrix}x&y&z\\u&v&w\end{bmatrix}

    \begin{bmatrix} 8&\text{-}1 \\ 4&\text{-}3\end{bmatrix} \cdot \begin{bmatrix} x&y&z \\ u&v&w \end{bmatrix}\;=\; \begin{bmatrix}49&4&24 \\ 27&12&12\end{bmatrix}


    \begin{bmatrix}8x-u & 8y-v & 8z-w \\ 4x-3u & 4y-3v & 4z-3w\end{bmatrix} \;=\;\begin{bmatrix}49&4&24\\27&12&12\end{bmatrix}


    \text{We have: }\:\begin{Bmatrix}8x-u &=& 49 \\  4x-3u &=& 27 \end{Bmatrix} \quad \begin{Bmatrix}8y-v &=& 4 \\ 4y-3v &=& 12 \end{Bmatrix} \quad \begin{Bmatrix}8z-w &=& 24 \\ 4z-3w &=& 12 \end{Bmatrix}


    \text{Solve the systems: }\:\begin{Bmatrix}x &=& 6 \\ u &=& \text{-}1 \end{Bmatrix} \quad \begin{Bmatrix}y &=& 0 \\ v &=& \text{-}4 \end{Bmatrix} \quad \begin{Bmatrix}z &=& 3 \\ w &=& 0 \end{Bmatrix}


    \text{Therefore: }\:\begin{bmatrix}x&y&z\\u&v&w\end{bmatrix} \;=\;\begin{bmatrix}6&0&3 \\ \text{-}1&\text{-}4&0\end{bmatrix}
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  8. #8
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    Re: Matrices

    Quote Originally Posted by JP2012 View Post
    Actually I figured it out, if you take the inverse of A and multiply it by the finished matrix you get the answer.
    In other words, you were asked to solve AX= B for X. It would have been nice if you had told us that!
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