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Consider the matrices, each of which is in upper row-echelon form $$ \begin{pmatrix} 0 & 1 &2\\ 0 & 2 & 3\\ 0 & 0 & 3 \\ \end{pmatrix} $$

................................................................................................................................................. $$ \begin{pmatrix} 1 & 0 &2\\ 0 & 0 & 3\\ 0 & 0 & 1 \\ \end{pmatrix} $$ I now two rules, to work out the rank, firstly is the number of steps, so the first matrix here should have a rank of 2, but also the I have seen a rule that states the rk is the number of none empty rows which in the first case would be one. So what do we do if we have all 0 its row-echelon form and rank?

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  • $\begingroup$ Rank is the number of "steps", or pivots. The other rule you see is incorrect, unless you apply it to a special matrix (such as $\begin{smallmatrix}I&0\\0&0\end{smallmatrix}$). $\endgroup$ – vadim123 Nov 7 '14 at 16:35
  • $\begingroup$ Forget these 'rules' you will not remember in the long term anyway. You know that the rank is invariant in the row-echelon transformation. What is the rank of the last matrix? $\endgroup$ – mookid Nov 7 '14 at 16:37
  • $\begingroup$ @vadim123 what do we mean by steps/pivots though? $\endgroup$ – user135842 Nov 7 '14 at 16:42
  • $\begingroup$ @Integrator I think so $\endgroup$ – user135842 Nov 7 '14 at 16:43
  • $\begingroup$ @Integrator Number of linearly independent columns, so it has to be 2, since in each case the (0,0,0) column can be formed from the linear combination of the others and the other two columns are linearly independent from each other. $\endgroup$ – user135842 Nov 7 '14 at 16:48
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In the second matrix, apply $R_3\rightarrow R_3-\frac{1}{3}R_2$ and we get its rank $2$ as it has two pivots. In the first matrix, apply $R_2\rightarrow R_2-2R_1$ and then apply $R_3\rightarrow +3R_2$; we get its rank 2 as we again get two pivots. The first non-zero entry of the row is called pivot; below which all entries in the matrix are zero. For null matrix (all zero entries); rank is zero.

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