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Thread: Consider 2 homogeneous systems

  1. #1
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    Consider 2 homogeneous systems

    Find all possible solutions to W1 and W2.
    Find the set of all vectors which are solutions to both systems and why it is a vector space.

    Please help.
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  2. #2
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    The augmented matrix for the first is

    $\displaystyle \begin{bmatrix}
    1 && 2 && 1 && 0 \\
    2 && 4 && 2 && 0 \\
    1 && 2 && 1 && 0 \\
    \end{bmatrix}
    $

    In reduced tow eschelon form you get...

    $\displaystyle \begin{bmatrix}
    1 && 2 && 1 && 0 \\
    0 && 0 && 0 && 0 \\
    0 && 0 && 0 && 0 \\
    \end{bmatrix}
    $

    so we have two "free" variables we get the solution

    $\displaystyle x=-2s-t$
    $\displaystyle y= s$
    $\displaystyle z=t $

    $\displaystyle v_=\begin{bmatrix}
    -2s-t \\
    s \\
    t
    \end{bmatrix}= s\begin{bmatrix}
    -2 \\
    1 \\
    0 \\
    \end{bmatrix}+ t\begin{bmatrix}
    -1 \\
    0 \\
    1 \\
    \end{bmatrix}$

    The last two vectors span the solution space of the first matrix.

    Try this proceedure with the second matrix and compare the solution spaces.

    I hope this helps

    Good luck.
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  3. #3
    Junior Member
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    thanks man.
    I got the following solution for second one.
    So is S[-2,1,0] the solution for both systems?
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