What is a solution to this matrix rank problem?

Question

Find the rank of a matrix depending on parameters r, s.

\begin{bmatrix} 1 &0 &0 \\ 2 &r-2 &2 \\ 0 &s-1 &r+2 \\ 0 &0 &3 \end{bmatrix}

My attempt was to interchange rows 3 and 4, and then using properties of row-echelon

rank = 3, if r=-2 and s=1

But this is only a partial solution.

Answer 1

The first and the last column are always going to be linearly independent, aren't they? So the rank here is at least 2. Any non-trivial linear combination of the first and last column will end up having non zero entries on the first or last rows; in other word, the second column never belongs to the space spanned by the two others columns, unless it is zero. When it happens ($s=1, r=2$) the rank is obviously 2, else it is 3.

Answer 2

If $r=2$ and $s=1$ then the matrix has rank 2, because the first and third columns will be linearly independent, while the second column is 0.

Otherwise, the matrix has rank 3, because by performing row and column operations, you can easily bring the matrix to have a 3x3 identity matrix in the top left corner.