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I have the matrix $$\pmatrix{-1 & -2 & 2 & -12 \\ 4 & 1 & -3 & 17 \\ 2 & -3 & 3 & -11}$$

I am trying to bring this to reduced row echelon form. Here are the operations I performed: $$R_2-2R_3 \\ R_3 + 2R_1 \\ R_3 + R_2 \\ - R_1 \\ R_2/7 \\ R_3/-2 \\ R_1-2R_2 \\ R_2+ (9/7) R_3$$ I arrived at $$\pmatrix{1 & 0 & 4/7 & 6/7 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -2}$$ I know the normal way to proceed is to create a leading 1 for the first row and then add and subtract multiples of the first row from other rows but why wont the above operations give me a correct RREF? Is there an inherent problem in the above operations? If so, what is it?

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  • $\begingroup$ Are you thinking that it is incorrect because the $4/7$ entry is not zero? $\endgroup$
    – wgrenard
    Oct 17, 2016 at 22:39
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    $\begingroup$ What you have is correct. You just need to perform one more elementary row operation before you have it in RREF. $\endgroup$
    – user137731
    Oct 17, 2016 at 22:46
  • $\begingroup$ @wgrenard This is a homework problem and it shows the 4/7 and 6/7 entries to be incorrect $\endgroup$
    – user140161
    Oct 17, 2016 at 22:48

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You're almost done you just have to do R1 - (4/7)R3 to get rid of the 4/7 in R1.

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