= Solution
Apply the <elementary row operations> from bottom to top: $R_4\gets R_4-R_3$, $R_3\gets R_3-R_2$, and $R_2\gets R_2-R_1$. This order ensures that each subtraction uses the original preceding row. The result is the <upper triangular matrix>
$$
T=\begin{pmatrix}a&a&a&a\\0&b-a&b-a&b-a\\0&0&c-b&c-b\\0&0&0&d-c\end{pmatrix}.
$$
Put $\delta_1=a$, $\delta_2=b-a$, $\delta_3=c-b$, $\delta_4=d-c$. Row $i$ is either zero if $\delta_i=0$, or has its first nonzero entry in column $i$. Thus, for arbitrary real parameters, a <row echelon form> is obtained by retaining the nonzero rows of $T$ in their existing order and moving every zero row to the bottom. This deals explicitly with vanishing parameters; an <upper triangular matrix> with an intermediate zero row is not yet a <row echelon form>.
The subtractions preserve the <determinant>, so the original <matrix> has
$$
\det M=a(b-a)(c-b)(d-c).
$$
A square <matrix> is invertible exactly when its <determinant> is nonzero. Consequently,
$$
\boxed{M\text{ invertible}\iff a\ne0,\quad b\ne a,\quad c\ne b,\quad d\ne c.}
$$
These conditions concern consecutive parameters: no ordering, positivity, or pairwise distinctness of all four is required. More generally the <rank of a matrix> here is the number of nonzero $\delta_i$, because the retained nonzero rows have distinct leading columns.
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