# Find matrix rank

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.

• How can you subtract both the second and the third row in the same time? Apr 26, 2019 at 9:45
• You subtract one row from another, leaving one of the rows. You don't subtract a third thing from both!
– Paul
Apr 26, 2019 at 9:50
• 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 Apr 26, 2019 at 10:26

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}