After row reduction, an inconsistent system looks like a row of zeros before the augmented matrix sign, but a nonzero entry after it. Such an equation can't hold, can it? So you have two choices here: either the system is consistent and has infinitely many solutions (but only one choice for ), or the system is inconsistent and has no solutions (but any other choice for is fine).
Yes, that is correct. But you are doing too much work. Your second step, adding a multiple of the second row to the first, is unecessary. Always work by columns, from left to right. Since there is already a "1" in the upper left, you don't need to do anything to the first row. Subtract twice the first row from the second and 4 times the first row from the third to get
Now since the first three numbers in the last two rows are the same, subracting the second row from the third will give "0 0 0" so the matrix is not "invertible"- there is not a unique solution, which is what dwsmith really meant to say. If , then so that the two last rows are identical. In that case, there are an infinite number of solutions so the system certainly is "consistent". If is any other number, there is NO solution which is what "inconsistent" means.