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Help solving a linear system

  1. Sep 24, 2012 #1
    Find the general solution to the following system of equations and indicate which variables are free and which are basic.

    x_1 + 4x_4 + 3 = x_2 + x_3
    x_1 + 3x_4 + 1 = \frac{1}{2}x_3
    x_1 + x_2 + 2x_4 = 1

    Putting it in augmented matrix form to start we have:



    1 -1 -1 4 | -3
    1 0 -1/2 3 | -1
    1 1 0 2 | 1

    Now performing the following fundamental row operations:

    R1<-->R2
    R2+R3-->R2
    -2R3+R2-->R2
    -R3+R1-->R3
    R2/-2
    R2+R3-->R2
    -3R3+R1-->R1

    And finally I end with the augmented matrix:

    1 0 -2 0 | 5
    0 1 0 0 | 0
    0 0 -1/2 1 |-2

    Can someone please tell me if I got the correct matrix at the end and if so how do I determine which variables are free and which are basic?

    Thank you.
     
  2. jcsd
  3. Sep 24, 2012 #2

    Simon Bridge

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    You were so close to getting the LaTeX - all you needed was a double-hash on each side of each line:

    ##x_1 + 4x_4 + 3 = x_2 + x_3##
    ##x_1 + 3x_4 + 1 = \frac{1}{2}x_3##
    ##x_1 + x_2 + 2x_4 = 1##

    The secret is to think about what you know at each stage...
    You have four unknowns and three equations - so what does that mean?

    Putting it in augmented matrix form to start we have:

    $$\left ( \begin{array}{cccc}
    1 & -1 & -1 & 4 \\
    2 & 0 & -1 & 6\\
    1 & 1 & 0 & 2
    \end{array}\right |
    \left . \begin{array}{c}
    -3\\-2\\1 \end{array}\right )$$
    ...I took the liberty of doubling row 2 to get rid of the annoying fraction.
    The row-reductions to echelon form follows more easily:
    R2 <--- R2-2R1; R3 <--- R3-R1
    ... does not lead to your matrix, no.
    You want to check my working though.

    I'll deal with the second question when you've done this.
     
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