The last step in row-reduction of the matrix is to turn all the leading entries into "1"s, as follows:

1 0 0 0 0 0 1 0 0 6 R2 0 0 1 0 4 R3 0 0 0 2 3 1 2 R2

1 0 0 0 0 0 1 0 0 6 0 0 1 0 4 0 0 0 1 3/2

Therefore, the solution is

x = 0, y = 6, z = 4, w = 2/3.

In vectort form, this is

(0, 6, 4, 2/3).