How are these matrices row equivalent?

According to Morris, a. O. (Linear Algebra-an introduction, second edition) the following two matrices are row equivalent. $$\begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix} \hspace{5ex} \begin{pmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 1 & -1 & 1 \end{pmatrix}$$ Reducing each matrix to echelon form, I get

$$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} \hspace{5ex} \begin{pmatrix} 3 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$ I am not able to reduce the matrices to exactly the same entries (except if I have to reduce each matrix into canonical form, in which case they both become identity matrices), so my contention was that they are row equivalent because a) they have the same rank and b) the pivot elements are in the exact same places. What is the correct answer? Correction to first row of first matrix is (1 0 1). Further correction to row reduced echelon form of the first matrix.

• – YoTengoUnLCD Nov 14 '15 at 21:55
• Have the answer/comment been of any help to you, Zilore? – Gerry Myerson Nov 16 '15 at 9:06

2 Answers

$$\pmatrix{1&0\cr0&0\cr}{\rm\ and\ }\pmatrix{1&1\cr0&0\cr}$$ have the same rank, and have pivot elements in exactly the same places, but they aren't row equivalent. You do have to go to what you call canonical form (what I call reduced row echelon form).

• Thanks @Gerry Myerson, I went through the wikipedia sites before asking because I was not satisfied with the content there. From you explanation the row canonical form results in the 3x3 identity matrix, and that was my question-should we go up to the identity matrix to make them row equivalent? – Zilore Mumba Nov 16 '15 at 21:06
• Yes. ${}{}{}{}$ – Gerry Myerson Nov 16 '15 at 22:19
• Thanks once more, now am clear. – Zilore Mumba Nov 17 '15 at 3:53
• If you found my answer helpful, you might want to "accept" it by clicking in the check mark next to it. – Gerry Myerson Nov 17 '15 at 5:31

two matrices are row equivalent means they have to has same row echelon form.