# What does it mean when the bottom row of a reduced row echelon form is all zeros with a 1 at the end?

I've just started my linear algebra course, and one question has me reduce an augmented matrix A of the form $$A=\begin{bmatrix} n_{11} & n_{12} & n_{13} & a\\ n_{21} & n_{22} & n_{23} & b\\ n_{31} & n_{32} & n_{33} & c \end{bmatrix}$$

But, after reducing this matrix to its reduced row echelon form: $$rref(A)=\begin{bmatrix} 1 & 0 & 0 & a-b\\ 0 & 1 & 0 & b\\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Since the last row ($$R_3$$) is all zeros save for the rightmost column of this augmented matrix, is this matrix consistent? As in, does the system of equations represented by this matrix have solutions? Because if $$R_3$$ was all zeros including the last column, then you could just ignore it, but if it equals 1, does that not imply $$0=1$$ which would be a contradiction? Would that contradiction mean that the matrix is not consistent? Or maybe it is only consistent with infinite solutions for a certain $$(a,b,c)$$?

Thanks

• It means that $0x+0y+0z=1$... which is clearly false Jun 7, 2021 at 20:09

Yes, it implies that $$0=1$$, which is nonsense, so the system is inconsistent. If you hypothetically had e.g. $$0=a^2$$ as the last row, with the freedom to choose $$a=0$$, then you would be able to say it has infinitely many solutions for $$a=0$$, but it's inconsistent for $$a\neq 0$$.