Question: Calculate the rank of the following matrices:
$A = \left( \begin{array}{cc} 1 & n \\ n & 1 \end{array} \right), n \in \mathbb{Z}$ and $B = \left( \begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array} \right)$, $x,y,z \in \mathbb{R}$.
So the way I understand rank($A$), is the number of pivots in an echelon form of $A$. To put $A$ into echelon form I would subtract $n$ times the first row from the second row: $A \sim \left( \begin{array}{cc} 1 & n \\ n & 1 \end{array} \right) \sim \left( \begin{array}{cc} 1 & n \\ 0 & 1 - n^{2} \end{array} \right) \Rightarrow $rank$(A) = 2$.
With $B$ I would have done pretty much the same thing, subtracting row 1 from both row 2 and row 3: $B \sim \left( \begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array} \right) \sim \left( \begin{array}{ccc} 1 & x & x^{2} \\ 0 & y - x & y^{2} - x^{2} \\ 0 & z - x & z^{2} - x^{2} \end{array} \right)$ (at this point I could multiply row 2 by $-(\frac{z-x}{y-x})$ and add it to row 3 which ends up being a long polynomial....) However, with both parts, I am pretty confident that it is not so simple and that I am missing the point of this exercise. Could somebody please help point me in the right direction?