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Thread: Linear system solutions using Gauss-Jordan elimination

  1. #1
    qaz
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    Linear system solutions using Gauss-Jordan elimination

    <br />
\left\{<br />
\begin{matrix}<br />
  ax - 3y + z = a\\<br />
  -x + 6y = -2\\<br />
  x + 5z = 4\\<br />
  \end{matrix}\right.<br />

    system matrix:
    <br />
\[<br />
A =<br />
\left[ \begin{array}{ccc|c}<br />
a & -3 & 1 & a\\<br />
-1 & 6 & 0 & -2\\<br />
1 & 0 & 5 & 4\end{array}\right]<br /> <br />

    I'm not sure where to start. Should I switch row 3 and row 1?
    Last edited by qaz; Jan 29th 2011 at 06:17 AM. Reason: spelling
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by qaz View Post
    <br />
\left\{<br />
\begin{matrix}<br />
  ax - 3y + z = a\\<br />
  -x + 6y = -2\\<br />
  x + 5z = 4\\<br />
  \end{matrix}\right.<br />

    system matrix:
    <br />
\[<br />
A =<br />
\left[ \begin{array}{ccc|c}<br />
a & -3 & 1 & a\\<br />
-1 & 6 & 0 & -2\\<br />
1 & 0 & 5 & 4\end{array}\right]<br /> <br />

    I'm not sure where to start. Should I switch the 3rd line with the 1st?
    What's the question? Are you supposed to solve for x, y, z in terms of a or find a value of a for which the system has a unique solution or ...?

    If you're required to solve for x, y, z in terms of a, you can indeed start by switching row 1 and row 3.
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  3. #3
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    Your first objective, in row reduction, is to get a "1" in the first column, first row. Yes, you can do that by swapping the first and third rows (which is a row operation). In general, as long as the number already is "first column, first row" is not 0, you can get a "1" there by dividing the entire row by that number (here, "a") but for this problem swapping rows is simpler.
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  4. #4
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    Quote Originally Posted by qaz View Post
    <br />
\left\{<br />
\begin{matrix}<br />
  ax - 3y + z = a\\<br />
  -x + 6y = -2\\<br />
  x + 5z = 4\\<br />
  \end{matrix}\right.<br />

    system matrix:
    <br />
\[<br />
A =<br />
\left[ \begin{array}{ccc|c}<br />
a & -3 & 1 & a\\<br />
-1 & 6 & 0 & -2\\<br />
1 & 0 & 5 & 4\end{array}\right]<br /> <br />

    I'm not sure where to start. Should I switch the 3rd line with the 1st?
    I would switch the 1st and 3rd, then add row 1 to 2, divide row 2 by 6, .....

    You keep going until you have an upper triangular matrix or you if you are feeling ambitious you can do the rref i.e. until you have the identity matrix.
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  5. #5
    qaz
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    Hello,
    So I did start by switching row 1 and row 3...

    <br />
\[<br />
\left[ \begin{array}{ccc|c}<br />
a & -3 & 1 & a\\<br />
-1 & 6 & 0 & -2\\<br />
1 & 0 & 5 & 4\end{array}\right]<br />


    \begin{array}{c}\\ R_3<->R_1\end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ -1&6&0&-2 \\ a&-3&1&a \end{array}\right]

    \begin{array}{c}\\ R_2+R_1\\R_3-aR_1\end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&-3&1-5a&-3a \end{array}\right]

    \begin{array}{c}\\ R_3+\frac{1}{2}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&-5a+7/2&-3a+1 \end{array}\right]

    \begin{array}{c}\\ \frac{1}{-5a+7/2}R_3 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&1&3/5a+2/7 \end{array}\right]

    \begin{array}{c}\\ \frac{1}{5}R_1-R_3\\\frac{1}{5}R_2-R_3 \end{array} \left[\begin{array}{ccc|c}1/5&0&0&3/5a-18/35 \\ 0&6/5&0&3/5a-4/35 \\ 0&0&1&3/5a+2/7 \end{array}\right]

    \begin{array}{c}\\ \frac{1}{5}R_1\\\frac{6}{5}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&0&3/25a-18/175 \\ 0&1&0&18/25a-24/175 \\ 0&0&1&3/5a+2/7 \end{array}\right]

    if a \ne \frac{5}{3}*-\frac{2}{7} than rank(A) = 3 = rank([A|B]) = 3 -> Possible,

    if a=\frac{5}{3}*-\frac{2}{7} than rank(A) = 3 > rank([A|B]) = 2 -> ??
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  6. #6
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    Quote Originally Posted by qaz View Post

    \begin{array}{c}\\ R_3+\frac{1}{2}R_2 \end{array} \left[\begin{array}{ccc|c}1&0&5&4 \\ 0&6&5&2 \\ 0&0&-5a+7/2&-3a+1 \end{array}\right]
    Assuming everything is correct. If a = 7/10, your matrix is inconsistent.

    If a = 1/2, then your solution is (x,y,z).
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