Are there values for $\displaystyle r$ and $\displaystyle s$ for which:

$\displaystyle \left(\begin{array}{ccc}1&0&0\\0&r-2&2\\0&s-1&r+2\\0&0&3\end{array}\right)$

has rank 1 or 2? And if so, find those values.

Well, it's obvious that if $\displaystyle r=2$ and $\displaystyle s=1$. It will have rank 2, and I can see that it can never have rank 1, but I am not sure how to exactly show this. I know in order to have rank 1, each row (or column) must be a scalar of every other column. So, would it be fair to say that $\displaystyle aC_{1}+bC_{2}+cC_{3}+dC_{4}=(0,0,0,0)$ for some non-trivial solution of $\displaystyle a,b,c,d$in order for that to be true. I think using the rows won't work in this situation since the rows can be thought of as vectors in $\displaystyle R^{3}$ and a set of 4 vectors in $\displaystyle R^{3}$ must be linearly dependent. I'm guessing if I can show that the a linear combination of the three column vectors can only equal a column of 0's with only the trivial solution, than that eventually leads to the fact that it is never possible to have any given row (or column) represented as a scalar multiple of all of the other rows or columns.

Any help would be appreciated.