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Row reduce the following matrix over $\mathbb Z_{3}$ to row-reduced echelon form:

$$M= \left[ {\begin{array}{cc} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 1 \\ \end{array} } \right]$$

I got the following answer:

\begin{array}{cc} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array}

but when using Mathematica to verify my answer, it gave me this answer:

\begin{array}{cc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}

Is my answer wrong, or are these equivalent?

UPDATE

So if I do R1 $\rightarrow$ R1 - R3 I get:

\begin{array}{cc} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array}

Is this still not in reduced row echelon form? If not, why?

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  • $\begingroup$ I updated my answer, is this still not in row reduced echelon? $\endgroup$ – user1282637 Sep 8 '14 at 1:46
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Your matrix is in 'row-echelon' form since below leading 1s there are 0s, but not necessarily above leading 1s. For reduced row echelon form you need to finish clearing the columns of all the leading 1s.

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  • $\begingroup$ So if I do R1 $\rightarrow$ R1 - R3 I get: \begin{array}{cc} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} Is this still not in reduced row echelon form? $\endgroup$ – user1282637 Sep 8 '14 at 1:44
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    $\begingroup$ Reduced row echelon form requires: (1) The leading nonzero entry of each row is a 1; (2) The leading 1 in a row is strictly to the right of the leading 1 in the row above it; (3) The column of any row-leading 1 consists of all 0s (other than the row leading 1); (4) Any all zero rows are at the bottom of the matrix. $\endgroup$ – paw88789 Sep 8 '14 at 1:46
  • $\begingroup$ I must have missed number (3) when computing/checking. Thanks for helping and not just giving answer! $\endgroup$ – user1282637 Sep 8 '14 at 1:49
  • $\begingroup$ @1282637 Glad to help! $\endgroup$ – paw88789 Sep 8 '14 at 1:51

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