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Thread: Gaussian Elimination

  1. #1
    Junior Member
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    Gaussian Elimination

    Hey, can anyone help to explain a way to do Gaussian elimination by hand (not calc). I seem to get incorrect answers by hand..


    For example, row operations for this matrix

    [[ 4 -2 1 20]
    [ 5 -4 -2 -28]
    [- 4 -5 -4 6]]

    Thanks )
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  2. #2
    Super Member Showcase_22's Avatar
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    $\displaystyle \begin{pmatrix}
    {4}&{-2}&{1}&{20}\\
    {5}&{-4}&{-2}&{-28}\\
    {-4}&{-5}&{-4}&{5}
    \end{pmatrix}$
    The first step I would do is:

    $\displaystyle \begin{pmatrix}
    {1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
    {5}&{-4}&{-2}&{-28}\\
    {-4}&{-5}&{-4}&{5}
    \end{pmatrix}$
    $\displaystyle R_1 \rightarrow \frac{1}{4}R_1$

    $\displaystyle \begin{pmatrix}
    {1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
    {0}&{-\frac{3}{2}}&{-\frac{13}{4}}&{-53}\\
    {0}&{-7}&{-3}&{25}
    \end{pmatrix}$
    $\displaystyle R_1\rightarrow R_1$
    $\displaystyle R_2 \rightarrow R_2-5R_1$
    $\displaystyle R_3 \rightarrow R_3+4R_1$

    $\displaystyle \begin{pmatrix}
    {1}&{-\frac{1}{2}}&{\frac{1}{4}}&{5}\\
    {0}&{1}&{\frac{13}{6}}&{\frac{106}{3}}\\
    {0}&{-7}&{-3}&{25}
    \end{pmatrix}$
    $\displaystyle R_2 \rightarrow -\frac{2}{3}R_2$

    $\displaystyle \begin{pmatrix}
    {1}&{0}&{\frac{4}{3}}&{\frac{68}{3}}\\
    {0}&{1}&{\frac{13}{6}}&{\frac{106}{3}}\\
    {0}&{0}&{\frac{73}{6}}&{\frac{817}{3}}
    \end{pmatrix}$
    $\displaystyle R_1 \rightarrow R_1+\frac{1}{2}R_2$
    $\displaystyle R_2 \rightarrow R_2$
    $\displaystyle R_3 \rightarrow R_3+7R_2$

    This is quite a complicated one! I'll leave the last steps for you to do. (Hint, it starts with $\displaystyle R_3 \rightarrow \frac{6}{73}R_3$.
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  3. #3
    MHF Contributor

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    $\displaystyle \begin{bmatrix}4 & -2 & 1 & 20 \\ 5 & -4 & -2 & -28 \\ -4 & -5 & -4 & 6\end{bmatrix}$

    We want to reduce the first three columns to the identity matrix and I recommend working one column at a time.

    We want "1 0 0" in the first column. Since there is a "4" in first column, first row, multiply the first row by 1/4. Since there is a "5" in first column, second row, subtract 5 times the new first row from the second row. Since there is a "-4" in the first column, third row, add 4 times the new first row to the third row:

    $\displaystyle \begin{bmatrix}1 & -\frac{1}{2} & \frac{1}{4} & 5 \\ 0 & -\frac{3}{2} & -\frac{13}{4} & -53 \\ 0 & -3 & -3 & 26\end{bmatrix}$

    In order to get a "1" in second column, second row, we must multiply the second column by -2/3. Then add 1/2 the new second row to the first row and add 3 times the new second row to the third row:
    $\displaystyle \begin{bmatrix}1 & 0 & \frac{4}{3} & \frac{68}{3} \\ 1 & 0 & \frac{13}{6} & \frac{106}{3} \\ 0 & 0 & \frac{4}{3} & 132\end{bmatrix}$

    Finally, multiply the third row by 3/4 to get a "1" in the third column, third row, subtract 4/3 the new third row from the first row and subtract 13/6 the new third row from the second row:
    $\displaystyle \begin{bmatrix}1 & 0 & 0 & \frac{226}{3} \\ 0 & 1 & 0 & -\frac{1075}{6} \\ 0 & 0 & 1 & 99\end{bmatrix}$
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