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I have a following matrix:

\begin{bmatrix}0&0&-1&5\\0&0&-3&8\\0&0&1&2\\\end{bmatrix}

After: Multiply first two rows by -1 and add first and third row together I get this:

\begin{bmatrix}0&0&-1&5\\0&0&-3&8\\0&0&0&7\\\end{bmatrix}

After: Subtract second row from the first multiplied by 3, I get this:

\begin{bmatrix}0&0&-1&5\\0&0&0&7\\0&0&0&7\\\end{bmatrix}

Therefore I could subtract second and third row and get only one non zero row, which would mean that the rank of the matrix is one. But the rank of this matrix is 2, according to the solution. What am I doing wrong? I've been several times over this.

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    $\begingroup$ How can you subtract both the second and the third row in the same time? $\endgroup$ – Hongyi Huang Apr 26 at 9:45
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    $\begingroup$ You subtract one row from another, leaving one of the rows. You don't subtract a third thing from both! $\endgroup$ – Paul Apr 26 at 9:50
  • $\begingroup$ While you subtract third row from the second in the last step, you will get $a_{34}$ as zero, and now you cannot get any row further zero $\endgroup$ – sawan kumawat Apr 26 at 10:26
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You are on the right track, but you have got a bit tangled up in the negatives. See below for the full steps to take to get to a rank of $2$

\begin{align} &\color{white}=\begin{pmatrix}0&0&-1&5\\ 0&0&-3&8\\ 0&0&1&2\end{pmatrix}\\\\ &= \begin{pmatrix}0&0&1&-5\\ 0&0&-3&8\\ 0&0&1&2\end{pmatrix}\tag{$R_1=-R_1$}\\\\ &=\begin{pmatrix}0&0&1&-5\\ 0&0&0&-7\\ 0&0&1&2\end{pmatrix}\tag{$R_2=R_2+3R_1$}\\\\ &=\begin{pmatrix}0&0&1&-5\\ 0&0&0&-7\\ 0&0&0&7\end{pmatrix}\tag{$R_3=R_3-R_1$}\\\\ &=\begin{pmatrix}0&0&1&-5\\ 0&0&0&1\\ 0&0&0&7\tag{$R_2=-\frac17 R_2$}\end{pmatrix}\\\\ &=\begin{pmatrix}0&0&1&-5\\ 0&0&0&1\\ 0&0&0&0\tag{$R_3=R_3 -7R_2$}\end{pmatrix} \end{align}

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Another answer has pointed out your error in row-reduction and shown the correct calculations, but you could’ve found the rank of the matrix with hardly any work at all: The rank of a matrix is the maximum number of linearly-independent columns and rows, i.e., row rank is always equal to column rank.

You can see at a glance that the rank of this matrix is at most two because two of the columns consist entirely of zeros. The two non-zero columns are obviously not scalar multiples, so they are linearly independent and thus the rank is exactly two.

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