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I'm trying to find the rref(A) where A = \begin{bmatrix}2&1&4&-4&11\\1&2&3&1&8\\1&1&2&-1&6\end{bmatrix}

I used row operations and came out with the following: \begin{bmatrix}1&0&0&-17&13\\0&1&0&8&-6\\0&0&1&-6&8\end{bmatrix}

However the Linear Algebra Toolkit came out with this: \begin{bmatrix}1&0&0&-3&3\\0&1&0&2&1\\0&0&1&0&1\end{bmatrix}

I read in other questions here that it isn't possible to have two different possible rrefs and I'm unclear as to what I've done incorrectly as the rows satisfy the requirements of a rref matrix. Can anyone identify what I've done incorrectly? I can submit my exact row operations if required. Thanks.

EDIT: Here's a list of my operations: $R_1-R_3,R_2-R_3, R_3-R_2, R_1-2R_2, R_3*R_2, R_2-R_3, R_1-2R_2$

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    $\begingroup$ The reduced row echelon form of a matrix is unique. Your matrix is in the right form, so you probably made a mistake along the way. Check your arithmetic. You might find a website that works out the rref showing you step bu step what it's doing. (But there can be more than one sequence of steps to get to the unique answer.). If you can't find an error in your calculation you can edit your question to show all your work and ask someone here to check it. $\endgroup$ – Ethan Bolker Apr 8 '17 at 13:44
  • $\begingroup$ The easiest way to get to the rref seems to be to first reverse the order of the rows. All elements on the diagonal turn out to be $1$ without having to do any divisions. $\endgroup$ – Fabio Somenzi Apr 8 '17 at 14:11
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The reduced row echelon form is unique. The are many paths to this form, but the destination is unique.


Let's look at the example problem. Use augmented reduction and clear the columns sequentially.

Column 1 $$ % E \left[ \begin{array}{rcc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|ccc} 2 & 1 & 4 & -4 & 11 & 1 & 0 & 0 \\ 1 & 2 & 3 & 1 & 8 & 0 & 1 & 0 \\ 1 & 1 & 2 & -1 & 6 & 0 & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % $$

Column 2 $$ % E \left[ \begin{array}{crc} 1 & -\frac{1}{3} & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % $$

Column 3 $$ % E \left[ \begin{array}{ccr} 1 & 0 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & -3 \\ \end{array} \right] % A \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % = % \left[ \begin{array}{cccrc|rrr} \boxed{1} & 0 & 0 & -3 & 3 & -1 & -2 & 5 \\ 0 & \boxed{1} & 0 & 2 & 1 & -1 & 0 & 2 \\ 0 & 0 & \boxed{1} & 0 & 1 & 1 & 1 & -3 \\ \end{array} \right] % $$

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