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Math Help - Matrices

  1. #1
    Junior Member
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    Post Matrices

    Can someone help me with this problem? Thanks!

    Solve the systems of equations. Use Gaussian elimination with back-substition or Gauss-Jordan Elimination.

    2x - y + 3z = 24
    2y - z = 14
    7x - 5y = 6
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  2. #2
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    Lexington, MA (USA)
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    Hello, cuteisa89!

    Solve the system of equations.
    2x - y + 3z = 24
    2y - z = 14
    7x - 5y = 6
    We have: . \begin{vmatrix}2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14 \\ 7 & \text{-}5 & 0 & | & 6\end{vmatrix}


    \begin{array}{c}\text{Move }R_3 \\ \text{to }R_1 \end{array}\;<br />
\begin{vmatrix}7 & \text{-}5 & 0 & | & 6 \\ 2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}


    \begin{array}{c}R_1-3R_2 \\ \\ \\ \end{array}\;\begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 2 & \text{-}1 & 3 & | & 24 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}


    \begin{array}{c} \\ \\ R_2-2R_1 \\ \end{array}\;<br />
\begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 0 & 3 & 21 & | & 156 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}


    \begin{array}{c} \\ \frac{1}{3}R_2 \\ \\ \end{array}\;<br />
\begin{vmatrix}1 & \text{-}2 & \text{-}9 & | & \text{-}66 \\ 0 & 1 & 7 & | & 52 \\ 0 & 2 & \text{-}1 & | & 14\end{vmatrix}


    \begin{array}{c}R_1+2R_2 \\ \\ R_3-2R_2\end{array}\;<br />
\begin{vmatrix}1 & 0 & 5 & | & 38 \\ 0 & 1 & 7 & | & 52 \\ 0 & 0 & \text{-}15 & | & \text{-}90\end{vmatrix}


    \begin{array}{c} \\ \\ -\frac{1}{15}R_3\end{array}\;<br />
\begin{vmatrix}1 & 0 & 5 & | & 38 \\ 0 & 1 & 7 & | & 52 \\ 0 & 0 & 1 & | & 6\end{vmatrix}


    \begin{array}{c}R_1-5R_3 \\ R_2-7R_3\\ \\ \end{array}\;<br />
\begin{vmatrix}1 & 0 & 0 & | & 8 \\ 0 & 1 & 0 & | & 10 \\ 0 & 0 & 1 & | & 6\end{vmatrix}


    Therefore: . (x,\,y,\,z)\;=\;(8,\,10,\,6)

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  3. #3
    Junior Member
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    Thumbs up

    Thank you so much!! I've been working on that problem for so long!
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