# Reduced row echelon form of an augmented matrix is not unique

I am given a sysetem of linear equations which after graphing, have no solution (the three lines intersect at different points). Now I am trying to prove this algebraically.

As an augmented matrix,

$$\begin{bmatrix} 1 & -1 & 3 \\ 1 & 1 & 1\\ 2 & 3 & 6\\ \end{bmatrix}$$

• $$R_{1}-R_{2} \Rightarrow R_{2}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 2 & 3 & 6\\ \end{bmatrix}$$

continue from here in $$(1)$$ or $$(2)$$

$$(1)$$

• $$R_{1}-\frac{1}{2}R_{3} \Rightarrow R_{3}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 0 & -\frac{5}{2} & 0\\ \end{bmatrix}$$

• $$-\frac{1}{2}R_{2} \Rightarrow R_{2}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & -\frac{5}{2} & 0\\ \end{bmatrix}$$

• $$\frac{5}{2}R_{2} + R_{3} \Rightarrow R_{3}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 0 & -\frac{5}{2}\\ \end{bmatrix}$$

Then in RREF

$$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1\\ 0 & 0 & -\frac{5}{2}\\ \end{bmatrix}$$

$$(2)$$

• $$-2R_{1}+R_{3} \Rightarrow R_{3}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 0 & 5 & 0\\ \end{bmatrix}$$

• $$-\frac{1}{2}R_{2} \Rightarrow R_{2}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 5 & 0\\ \end{bmatrix}$$

• $$-5R_{2} + R_{3} \Rightarrow R_{3}$$

$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 0 & 5\\ \end{bmatrix}$$

Then in RREF

$$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1\\ 0 & 0 & 5\\ \end{bmatrix}$$

which is different from the RREF in $$(1)$$

Can someone explain why I end up with a different RREF? I thought all RREF are unique, but clearly not in this case. Of course as mentioned earlier, the system has no solutions and both augmented matrices show this but their RREF's are not unique still.

• I'm not sure if RREF are only unique for coefficient matrices instead of augmented matrices. Or if RREF are only unique for consistent systems Sep 16, 2020 at 0:43