# Can a system of linear equations be inconsistent even if the last column is not a pivot column.

I have a simple question which is troubling me.

I have seen this theorem in Linear Algebra which I quote here :

The Row Echelon Form of an Inconsistent System: An augmented matrix corresponds to an inconsistent system of equations if and only if the last column (i.e., the augmented column) is a pivot column.

Suppose I have a RREF augmented matrix with the last row containing $$0s$$ in the coefficient and 1 in the augmented column ( I mean $$0 0 ... 0|1$$ as the last row). Surely, the presence of this row would make the whole system inconsistent, as it would imply that $$0=1$$, which is not possible.

But, what I don't understand is that for this to happen, why is there a need for all the other entries of the last column to be zero. This doubt stems from the definition of pivot column which says it is a column containing pivot, and hence we know that all the entries in a column containing pivot must be zero.

But I don't see why other entries should be zero. In all the examples of inconsistent system RREF matrices also the last column is always having all the entries except the pivot to be zero.

I hope I made myself clear. I would be grateful to you if you can help me with this.

Thanks.

Within the augmented matrix, the $$1$$ is the pivot entry of the row $$[0\ \cdots \ 0\mid 1]$$ since it is the first non-zero entry of this row. By the definition of RREF, this pivot entry must be the only non-zero entry in its column, which means that all other entries of the last column must be zero.
If you have a matrix whose last row is $$[0\ \cdots \ 0\mid 1]$$, then it is indeed true that the associated system must be inconsistent. However, if there are any other non-zero entries in the last column, then the matrix is not in RREF.