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Thread: Matrix- using row operations

  1. December 2nd 2009, 04:55 PM #1
    kelsikels
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    Matrix- using row operations

    1.) 7x + 5y - z = 94
    x + 5y + 2z = 52
    2x + y + z = 29

    Help ASAP please! thanks

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  2. December 2nd 2009, 05:55 PM #2
    Haversine
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    Quote Originally Posted by kelsikels View Post
    1.) 7x + 5y - z = 94
    x + 5y + 2z = 52
    2x + y + z = 29

    Help ASAP please! thanks

    Set up:

    <br /> \left(\begin{array}{ccc|c}<br /> 7 & 5 & -1 & 94 \\<br /> 1 & 5 & 2 & 52 \\<br /> 2 & 1 & 1 & 29<br /> \end{array}\right)<br />

    Move the 2nd row to the top row. You'll see why.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 7 & 5 & -1 & 94 \\<br /> 2 & 1 & 1 & 29<br /> \end{array}\right)<br />

    Take the 2nd row: subtract 7x the top row from each entry.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & -30 & -15 & -270 \\<br /> 2 & 1 & 1 & 29<br /> \end{array}\right)<br />

    Divide all entries in 2nd row by -15.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & 2 & 1 & 18 \\<br /> 2 & 1 & 1 & 29<br /> \end{array}\right)<br />

    Third row: subtract 2x the top row from each entry.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & 2 & 1 & 18 \\<br /> 0 & -9 & -3 & -75<br /> \end{array}\right)<br />

    Divide each entry in the 3rd row by -3.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & 2 & 1 & 18 \\<br /> 0 & 3 & 1 & 25<br /> \end{array}\right)<br />

    Third row: subtract 3/2x of the 2nd row.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & 2 & 1 & 18 \\<br /> 0 & 0 & -\frac{1}{2} & -2<br /> \end{array}\right)<br />

    Multiply the entries in the 3rd row by -2.

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 5 & 2 & 52 \\<br /> 0 & 2 & 1 & 18 \\<br /> 0 & 0 & 1 & 4<br /> \end{array}\right)<br />

    This can be solved by back substitution, now. If you want to practice your row skills, find the answer by back substitution, then check yourself: find the answer directly by reducing it to this form:

    <br /> \left(\begin{array}{ccc|c}<br /> 1 & 0& 0 & x \\<br /> 0 & 1 & 0 & y \\<br /> 0 & 0 & 1 & z<br /> \end{array}\right)<br />

    You might also do the steps above yourself and check my work. Maybe I tossed in a monkey wrench somewhere along the way ...?
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