ok, when we solve a system of linear equations, which we can write in short-hand form as AX = B, one thing that is often done is to look at the homogeneous system AX = 0 first. why? because knowing the dimension of the null space {X in R^n: AX = 0}, tells us some qualitative things about the solution set {X in R^n: AX = B}.

in particular, if X1 is any specific solution to AX = B, and X2 is in the null space, then A(X1 + X2) = A(X1) + A(X2) = B + 0 = B. so if we want to find ALL the solutions to AX = B, we need to find ONE solution for AX = B, and ALL the solutions to AX = 0. often this is summed up as "general solution = particular solution + homogeneous solutions".

when the rank of the matrix A (i am presuming this is what they mean by "A' has d steps") is less than the dimension of the co-domain of A (R^m), there are going to be vectors B in R^m that aren't in the image of A. when the rank is less the the dimension of the domain of A (R^n) (often expressed as "fewer equations than unknowns", although this is not very accurate), the "n-d parameters" are essentially what we need to specify to include all the solutions that the null space of A gives us (since dim(null(A)) = n-d), over and above any particular solution we find.

for example, consider the following system:

x + 3z = 1

y + 4z = 2

the matrix A =

[1 0 3]

[0 1 4]

which is already in reduced row-echelon form. the null space of A is all vectors in R^3 of the form: (-3t,-4t,t) = t(-3,-4,1), which has the basis {(-3,-4,1)}. now to find a particular solution, we could let z = 0 (which is convenient), so (1,2,0) is a particular solution. then the general solution is: (1,2,0) + t(-3,-4,1).

in geometric terms, the solution set (of AX = B) is just the line that makes up the null space of A, "translated" by a particular vector in the solution set.