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Math Help - Another matrices word problem :(

  1. #1
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    Another matrices word problem :(

    hey guys, how to solve this word problem using gauss elimination method. u know using matrices and etc :P

    the problem is :
    A farmer can buy four types of plant food. Each barrel of mix A contains 30 pound of phosphoric acid, 50 pounds oof mitrogen and 20 pounds of potash; each barrel of mix B contains 30 pounds of phosphoric acid, 75 pounds of nitrogen . and 30 pounds of potash; each barrel of mix c contains 30 pounds of phosphoric acid, 25 pounds of nitrogen and 30 pounds of potash; each barrel of mix D contains 60 pounds of phosphoric acid, 25 pounds of nitrogen, and 50 pounds of potash. Sol tests indicate that a particular field needs 900 pounds of phosphoric acid. 700 pounds of nitrogen . and 820 pounds of potash. how many barrel of each type of food should the farmer mix together to supply the necessary nutrients for the field ?
    Let X1 be the number of barrels of Mix A, X2 be the number of barrels of Mix B , and X3 be the number of barrels of Mix C, and X4 be the number of barrels of Mix D.

    i just need help with the equations to start with.
    and please tell me how do u decide those equations also ? ty alot for everybody`s time.
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  2. #2
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    Re: Another matrices word problem :(

    {\bf a}=\left[\begin{array}{c} 30\\50 \\ 20 \end{array}\right],\,\, {\bf b}=\left[\begin{array}{c}30 \\75 \\30  \end{array}\right],\,\, {\bf c}=\left[\begin{array}{c} 30\\ 25\\ 30 \end{array}\right],\,\,{\bf d}=\left[\begin{array}{c} 60\\25 \\50  \end{array}\right],\,\,{\bf r}=\left[\begin{array}{c} 900\\700 \\820  \end{array}\right]

    x_1{\bf a}+x_2{\bf b}+x_3{\bf c}+x_4{\bf d}={\bf r}

    [{\bf a} \,\,{\bf b}\,\,{\bf c}\,\,{\bf d}]\cdot \left[\begin{array}{c} x_1\\x_2 \\x_3\\x_4  \end{array}\right]={\bf r}

    \left[\begin{array}{cccc} 30 & 30 & 30 & 60\\ 50 & 75 & 25 & 25 \\ 20 & 30 & 30 & 50  \end{array}\right]\cdot \left[\begin{array}{c} x_1\\x_2 \\x_3\\x_4  \end{array}\right]=\left[\begin{array}{c} 900\\700 \\820  \end{array}\right]

    Don't overlook the fact that every 4 vectors in 3 dimensional space are linearly dependent. It seems that b=2c+a-d.
    Thanks from romeroroma
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  3. #3
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    Re: Another matrices word problem :(

    Hello, romeroroma!

    A farmer can buy four types of plant food.
    Each barrel of mix A contains 30 pounds of phosphoric acid, 50 pounds of mitrogen and 20 pounds of potash.
    Each barrel of mix B contains 30 pounds of phosphoric acid, 75 pounds of nitrogen, and 30 pounds of potash.
    Each barrel of mix C contains 30 pounds of phosphoric acid, 25 pounds of nitrogen and 30 pounds of potash.
    Each barrel of mix D contains 60 pounds of phosphoric acid, 25 pounds of nitrogen, and 50 pounds of potash.
    Soil tests indicate that a particular field needs 900 pounds of phosphoric acid. 700 pounds of nitrogen.
    . . and 820 pounds of potash.
    How many barrels of each type of mix should the farmer combine to supply the necessary nutrients for the field?

    Let a = barrels of mix A, b = barrels of Mix B, c = barrels of Mix C, d = barrels of Mix D.

    First, let's tabulate the information . . .

    . .  \begin{array}{|c|c|c|c|}\hline & \text{phos.} & \text{nitr.} & \text{pot.} \\ \hline A & 30 & 50 & 20 \\ B & 30 & 75 & 30 \\ C & 30 & 25 & 30 \\ D & 60 & 75 & 50 \\ \hline \text{Total} & 900 & 780 & 820 \\ \hline \end{array}

    We have: . \begin{Bmatrix}30a + 30b + 30c + 60d &=& 900 \\ 50a + 75b + 75c + 25d &=& 780 \\ 20a + 30b + 30c + 50c &=& 820 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix}a + b + c + 2d &=& 30 \\ 2a + b + c + d &=& \frac{156}{5} \\ 2a + 3b + 3c + 5d &=& 82 \end{Bmatrix}

    With four variables and only three equations, we cannot expect a unique solution.


    We have: . \left|\begin{array}{cccc|c}1&1&1&2&30 \\ 2&1&1&1&\frac{156}{5} \\ 2&3&3&5 & 82 \end{array}\right|


    \begin{array}{c} \\ R_2-2R_1 \\ R_3-2R_1 \end{array}  \left|\begin{array}{cccc|c} 1&1&1&2&30 \\ 0&\text{-}1&\text{-}1&\text{-}3&\text{-}\frac{144}{5} \\ 0 &1&1&1&22 \end{array}\right|


    . . \begin{array}{c}R_1+R_2 \\ \text{-}1\!\cdot\!R_2 \\ R_3+R_2\end{array}\left|\begin{array}{cccc|c} 1&0&0&\text{-}1 & \frac{6}{5} \\ 0&1&1&3&\frac{144}{5} \\ 0&0&0&\text{-}2 & \text{-}\frac{34}{5} \end{array}\right|


    \begin{array}{c}R_1+R_3 \\ R_2-3R_3 \\ \end{array}\left|\begin{array}{cccc|c} 1&0&0&0 & \frac{23}{5} \\ 0&1&1&0 & \frac{93}{5} \\ 0&0&0&1&\frac{17}{5} \end{array}\right|


    \begin{array}{c}\\ \\ \text{-}\frac{1}{2}R_3 \end{array} \left|\begin{array}{cccc|c} 1&0&0&\text{-}1 & \frac{6}{5} \\ 0&1&1&3&\frac{144}{5} \\ 0&0&0&1 & \frac{17}{5} \end{array}\right|


    \begin{array}{c} R_1+R_3 \\ R_2 - 3R_3 \\ \end{array}\left|\begin{array}{cccc|c} 1&0&0&0&\frac{23}{5} \\ 0&1&1&0 & \frac{93}{5} \\ 0&0&0&1 & \frac{17}{5}\end{array}\right|


    \text{Therefore: }\:\begin{Bmatrix}a \;=\;\frac{23}{5} \\ b+c \;=\;\frac{93}{5} \\ d \;=\;\frac{17}{5} \end{Bmatrix}
    Thanks from romeroroma
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  4. #4
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    Re: Another matrices word problem :(

    oh. 1 more thing ..
    Last edited by romeroroma; May 5th 2012 at 02:13 AM.
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  5. #5
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    Re: Another matrices word problem :(

    All the solutions u gave me is great and true. but am still having the problem of choosing which one of these choices. here is a screen shot of my Work Math Book.

    Another matrices word problem :(-untitled.png

    so i dunno which 1 to choose and if i choose B what to fill in the boxes ? still uncomplete solution to me

    and ofc If the number of equations is less than the number of unknowns we will have either No solution or Infinite number of solutions and never a unique solution.
    Last edited by romeroroma; May 5th 2012 at 02:30 AM.
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