Originally Posted by
HallsofIvy That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.