first of all, you didn't "finish" row-reduction, because you only reduced your matrix to upper-triangular form (and not row echelon, or reduced row echelon form, which makes the solutions more transparent).

but look, your last row means that:

0x + 0y - 3z = -3 that is:

-3z = -3

z = 1 <--not a parameter, a VALUE.

your second row means:

0x + 2y + 6z = 2 that is:

2y + 6z = 2

but we already know that z = 1! so this becomes:

2y + 6 = 2

2y = -4

y = -2 <---again, not a parameter, an actual number.

finally, your top row means:

x + 4y + 11z = 7

since y = -2, ans z = 1, this is:

x - 8 + 11 = 7

x + 3 = 7

x = 4

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the "other" question you ask, how do you know "when to use parameters?" depends on the RANK of your original matrix.

row-reduction is one way we find out the rank, after row-reduction, it is the number of non-zero rows remaining. in this case, the rank of the original matrix is 3.

this means several things, all of which are equivalent:

1) the matrix A in your original problem is invertible

2) the matrix A is non-singular (the null space is ONLY the 0-vector (0,0,0)) that is A(x,y,z) = (0,0,0) implies x = y = z = 0.

3) the column space of A has dimension 3 (it is thus ALL of R^{3}).

4) the columns of A are linearly independent

5) the equation A(x,y,z) = (7,9,4) has a UNIQUE solution

6) the determinant of A is non-zero

7) we can use Cramer's Rule to solve the system of equations