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Is this the correct way to compute the row echelon form?

  1. Sep 29, 2015 #1
    This is actually a pretty simple thing, but the ref(A) that I compute on paper is different from the ref(A) that my TI-89 gives me.

    Compute ref(A) where A = [itex]
    \begin{bmatrix}
    1 & 2\\
    3 & 8
    \end{bmatrix}
    [/itex]

    [tex]\\ \begin{bmatrix}1 & 2\\ 3 & 8\end{bmatrix} \ r_2 \rightarrow r_2 - 3 \times r_1 \\ \\ \begin{bmatrix}1 & 2\\ 0 & 2 \end{bmatrix} \ r_2 \rightarrow \frac{1}{2} \times r_2 \\ \\ \begin{bmatrix}1 & 2\\ 0 & 1 \end{bmatrix}
    [/tex]

    Now I would have thought that this last matrix, A = [itex]
    \begin{bmatrix}
    1 & 2\\
    0 & 1
    \end{bmatrix}
    [/itex] would be the ref(A).

    But my TI-89 gives ref(A) = [itex]
    \begin{bmatrix}
    1 & \frac{8}{3}\\
    0 & 1
    \end{bmatrix}
    [/itex] and this is not the rref(A), the rref(A) is just the 2x2 identity matrix.
     
  2. jcsd
  3. Sep 29, 2015 #2

    Mark44

    Staff: Mentor

    Yes, I agree.
    I'm guessing that your calculator switched the two rows, and then did row reduction. If you start with this matrix --
    \begin{bmatrix}
    3 & 8\\
    1 & 2 \end{bmatrix}
    -- row reduction gives you the matrix the calculator shows.
     
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