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Math Help - reduce row echelon question

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    reduce row echelon question

    Let the reduced row echelon form of A be

    \left(\begin{array}{ccccc} 1 & 0 & 2 & 0 & -2\\ \\0 & 1 & -5 & 0 & -3\\ \\0 & 0 & 0 & 1 & 6\end{array}\right)

    Determini A if the first second and fourth columns of A are

    \left(\begin{array}{c} 1 \\ \\ - 1 \\ \\3 \end{array}\right), \left(\begin{array}{c} 0 \\ \\ - 1 \\ \\1 \end{array}\right) and \left(\begin{array}{c} 1 \\ \\ - 2 \\ \\0 \end{array}\right)

    would this just mean that you substitute the first second and fourth and perform row operations?
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    Quote Originally Posted by lllll View Post
    Let the reduced row echelon form of A be

    \left(\begin{array}{ccccc} 1 & 0 & 2 & 0 & -2\\ \\0 & 1 & -5 & 0 & -3\\ \\0 & 0 & 0 & 1 & 6\end{array}\right)

    Determini A if the first second and fourth columns of A are

    \left(\begin{array}{c} 1 \\ \\ - 1 \\ \\3 \end{array}\right), \left(\begin{array}{c} 0 \\ \\ - 1 \\ \\1 \end{array}\right) and \left(\begin{array}{c} 1 \\ \\ - 2 \\ \\0 \end{array}\right)

    would this just mean that you substitute the first second and fourth and perform row operations?
    In the Matrix A we need to express the columns without 1 as a liner combination of the colums with ones.

    for example

    c_3=2 \cdot c_1-5 \cdot c_2 and

    c_5=-2 \cdot c_1-3 \cdot c_2 + 6 \cdot c_4

    These are called the dependency equations

    You can find the original column vectors using the same equations.

    so

    c_3= 2 \cdot \begin{bmatrix}1 \\ -1 \\ 3 \end{bmatrix}<br />
-5 \cdot \begin{bmatrix}0 \\ -1 \\ 1 \end{bmatrix} = <br />
\begin{bmatrix}2 \\ 3 \\ 1 \end{bmatrix}
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