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Thread: Basic matrix problem with finding general solutions

  1. #1
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    Question Basic matrix problem with finding general solutions

    So basically, I have these two: Imgur: The most awesome images on the Internet examples in my book and i have no idea how they came to that conclusion. I'm obviously missing something fundamental here but I cannot see how you get those columns from those systems.
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  2. #2
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    Re: Basic matrix problem with finding general solutions

    The problem is "Find the general solution of the system x_1+ 2x_2- 3x_3= 5".

    They write the solution as \vec{x}= \begin{bmatrix}5 \\ 0 \\ 0\end{bmatrix}+ s\begin{bmatrix}-2 \\ 1 \\ 0 \end{bmatrix}+ t\begin{bmatrix}3 \\ 0 \\ 1 \end{bmatrix}

    Clearly, they are writing x_1, x_2, x_3 as components of the vector \vec{x}= \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}. Since x_1+ 2x_2- 3x_3= 5 we have immediately x_1= 5- 2x_2+ 3x_3.

    So \vec{x}= \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}= \begin{bmatrix}5- 2x_2+ 3x_3 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}+ \begin{bmatrix}-2x_2 \\ x_2 \\ 0 \end{bmatrix}+ \begin{bmatrix}3x_3 \\ 0 \\ x_3\end{bmatrix} = \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}+ x_2\begin{bmatrix}-2 \\ 1 \\ 0 \end{bmatrix}+ x_3\begin{bmatrix}3 \\ 0 \\ 1\end{bmatrix}.

    They have chosen to replace " x_2" and " x_3" with "s" and "t".

    The second problem is exactly the same thing except that they have "= 0" rather than "= 5" so there is no \begin{bmatrix}5 \\ 0 \\ 0 \end{bmatrix}. They are using the phrase "solution space" rather than "solution set" because, since this is a plane that includes the origin, it forms a vector space where the previous solution set does not. (Often a plane that does not include the origin is called a "linear manifold".)
    Last edited by HallsofIvy; Jan 10th 2017 at 09:21 AM.
    Thanks from Joppe
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