fastest way to identify "singular row" of a matrix
This is not an answer, but a response to this comment.
Assume that $A$ is the matrix which columns are the rows that we want to test for linear independence. I am putting them are columns for no other reason than personal comfort.
Doing the row transformations required in Gaussian elimination correspond to multiplying $A$ from the left by some elementary matrices. Assume that $T$ is the product of all elementary matrices corresponding to the row transformations needed to put $A$ in echelon form. So, $TA$ is is echelon form.
If each column has an echelon then the system $TAx=0$, with $x$ a column vector of the right size, has only the solution $x=0$. Since $T$ is invertible, we have that $Ax=0$ has only the solution $x=0$. Since $Ax$ is nothing else than a linear combination of the columns of $A$ using the components of $x$ as coefficients, what we got is that the columns of $A$ are linearly independent.
Now assume that some columns of $TA$ don't have echelons. Let's call these the dependent columns. You can give any values you want to the components of $x$ corresponding to the positions of the dependent columns and then solve the resulting system $TAx=0$ for the remaining components, by backward substitution. In particular we can put non-zero values to the components of $x$ corresponding to the dependent columns of $TA$. This gives you a solution of $TAx=0$. Since $T$ is invertible, it is also a solution of $Ax=0$. For this $x$ the expression $Ax$ is a linear combination of the columns of $A$ that is equal to zero and the coefficients of the dependent columns are non-zero. So, the equation $Ax=0$ expresses each of the dependent columns in terms of all the other columns.
So, the dependent columns are linearly dependent of the columns with echelon in them (independent columns).