# 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

Any RREF must have pivot 1 in each row. Your matrices do not satisfy this condition (look at row 3). So these are not RREFs. RREF of any matrix is unique. It is not a completely trivial statement, but it is a fact.

• My example are augmented matrices, that is in this case, there are more equations(3) then unknowns(2) Sep 16, 2020 at 0:45
• It does not matter. The RREF of every matrix, by definition, has pivot 1 in every row. Does not matter where the matrix comes from. Sep 16, 2020 at 0:47
• ? I thought there's only 4 conditions for RREF: 1. first nonzero entry in any nonzero row is 1, 2. zero rows at the bottom, 3. leading 1's to the right in lower than upper rows, 4. zeros below and above each leading 1 of each pivot column. My matrix satisfies all these Sep 16, 2020 at 0:49
• Yes, that is true. Your matrices do not satisfy condition 1. Sep 16, 2020 at 0:50
• There are only two pivot columns. The third column is the column representing the constant on the right side of the linear equation. The number in the last column last row is not supposed to be a leading 1 Sep 16, 2020 at 0:52