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I am supposed to row reduce a matrix to reduced row echelon form. $$ \begin{bmatrix} 1 & 2 & 4 & 8\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

I have tried the following: 4 times row 2 - row 1 gives the following matrix: $$ \begin{bmatrix} 1 & 2 & 0 & 8\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

I checked the answer in the back of the textbook I am using and it says that the answer is $$ \begin{bmatrix} 1 & 2 & 0 & -8\\ 0 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{bmatrix} $$ I see that this answer was obtained by using row 1 - 4 times row 2. Are both of these matrices equal? I thought that there could only be one unique matrix in reduced echelon form.

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It is true that the reduced row echelon form of a matrix is unique. Using this fact and checking some arithmetic with your calculation should help you.

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  • $\begingroup$ I can't seem to figure out what I am doing wrong. I have gone over my arithmetic many times and it seems correct. $\endgroup$ – Steven Jun 29 '14 at 1:05
  • $\begingroup$ You are adding negative copies of row 2 to row 1 to clear out the 4 in row one. What effect does this row operation have on the rest of row 1? $\endgroup$ – James Jun 29 '14 at 1:07
  • $\begingroup$ That was what I observed according to the answer in the back of the book. What I originally did was take 4 times row 2 and subtract row 1, then replace row 1 with that. Doing so would give me a positive 8 and not negative. $\endgroup$ – Steven Jun 29 '14 at 1:09
  • $\begingroup$ Doing that would not lead to reduced row echelon form. I would suggest following the Gaussian elimination steps very carefully to get a good feel for basic matrix operations. $\endgroup$ – James Jun 29 '14 at 1:18
  • $\begingroup$ Could you explain why it would not lead to reduced row echelon form? The only difference between the matrices is a negative sign in front of the 8. $\endgroup$ – Steven Jun 29 '14 at 1:23

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