Given the matrix A defined as A = \begin{bmatrix} 2 & -4 & 4 & -2 \\ 6 & a-2 & 7 & a-6 \\ -1 & 2-b & 8 & -b+1 \\ \end{bmatrix}
where (a) and (b) are real constants, I am interested in finding the rank and nullity of (A). Applying the normal Reduced Row Echelon Form (RREF) method, I obtained a matrix R equivalent to R = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}
yielding a rank of 3 and a nullity of 1 for (A).
My question is whether this approach directly applies despite the presence of parameters (a) and (b). Specifically, is it necessary to consider different cases based on the values of (a) and (b) (e.g., (a = 0), (b = 0), (a ≠ 0), (b ≠ 0)) to accurately determine the rank and nullity, or can the RREF result be used universally for any real values of (a) and (b)?