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Thread: How to choose free variable?

  1. #1
    Oct 2010

    How to choose free variable?

    I thought choosing the free variable was simple (and it probably is), but I'm having difficulty on this one.

    The RREF of the matrix is:

    $\displaystyle $\begin{array}{ccccc}
    1 & 0 & 0 & \frac{2}{5} & 0\\
    0 & 1 & 0 & \frac{-3}{5} & 0\\
    0 & 0 & 1 & \frac{-1}{5} & 0\end{array}$$

    Now, I thought the free variables were supposed to be those that aren't associated with a leading one in a reduced form of the matrix. However,the answer for this problem has C3 as the free variable. C3 corresponds to a leading one! How is this?
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Technically, you can choose any variable to be a parameter as long as its value is not completely determined. I tend to go from the right-hand-side of the matrix towards the left-hand-side, as, it appears, you do also. Let's take the following ref augmented matrix as an example:

    $\displaystyle \left[\begin{array}{ccc|c}
    1 &0 &2 &3\\
    0 &1 &5 &7

    Method one for back substitution:

    Let $\displaystyle z=t.$ Then we have the equation $\displaystyle y+5t=7,$ or $\displaystyle y=7-5t.$ We also have the equation $\displaystyle x+2t=3,$ or $\displaystyle x=3-2t.$ The solution is then

    $\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix }3\\7\\0\end{bmatrix}+t\begin{bmatrix}-2\\-5\\1\end{bmatrix}.$

    Method two for back substitution:

    Let $\displaystyle y=s.$ Then we have the equation $\displaystyle s+5z=7,$ or $\displaystyle 5z=7-s,$ or $\displaystyle z=7/5-s/5.$ We also have the equation $\displaystyle x+2(7/5-s/5)=3,$ or $\displaystyle x+14/5-2s/5=3$, or $\displaystyle x=1/5+2s/5.$ The solution is then

    $\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix }1/5\\0\\7/5\end{bmatrix}+s\begin{bmatrix}2/5\\1\\-1/5\end{bmatrix}.$

    Question: are these the same solution?

    Well, both solutions are the equation of a straight line. So, are they both describing the same straight line? Well, if they're parallel and have at least one point in common, then they are the same line. That the lines are parallel you can tell because the direction vectors (the vectors multiplying the parameters $\displaystyle t$ or $\displaystyle s$) are multiples of each other. I can multiply the $\displaystyle t$ direction vector by $\displaystyle -1/5$ to get the $\displaystyle s$ direction vector. Do they have a point in common? Well, clearly, the point $\displaystyle (3,7,0)$ is on the $\displaystyle t$ line, which point I got simply by setting $\displaystyle t = 0$. Is that point also on the $\displaystyle s$ line? We set

    $\displaystyle \begin{bmatrix}3\\7\\0\end{bmatrix}=\begin{bmatrix }1/5\\0\\7/5\end{bmatrix}+s\begin{bmatrix}2/5\\1\\-1/5\end{bmatrix}.$

    The second component tells us that $\displaystyle s=7,$ and you can tell by inspection that that value for $\displaystyle s$ also works for the first and third components. Hence, $\displaystyle (3,7,0)$ is on both lines, and the lines are the same.

    Therefore, the solutions are the same.

    What it amounts to is a re-parametrization of the solution. The actual vectors in the solution space are not going to change one way or the other.

    Does all that make sense?
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