# Issue understanding the difference between reduced row echelon form on a coefficient matrix and on an augmented matrix

I have a problem understanding something about matrices and the difference between a coefficient matrix and an augmented matrix. One theorem in a book I'm reading states:

Suppose thate $A\mathbf{x = b}$ is a sytem of $m$ linear equations in $n$ variables. The system is consistent iff the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $\left[\begin{array}{c|c}A&b\end{array}\right]$.

With the following definition for rank of a matrix:

The rank of a matrix $M$ is the number of leading 1's in the reduced row echelon form that is row equivalent to $M$.

What is confusing me about this statement is the difference between the rank of the coefficient matrix $A$ and the augmented matrix $\left[\begin{array}{c|c}A&b\end{array}\right]$. The definition of rank requires knowing the reduced row echelon form of a matrix, but how can one find the reduced row echelon form of the coefficient matrix $A$? Wouldn't you always need the augmented matrix? And once you've transformed $A$ into reduced row echelon form, why would it ever be different from the reduced row echelon form of $\left[\begin{array}{c|c}A&b\end{array}\right]$?

In short, row reduced echelon form(RREF) of a matrix $A$ is such that

i) Every leading entry is 1

ii) Any nonzero rows are above zero rows

iii) any leading entry is strictly to the right of any leading entries above that row

iv) any other entry in a column containing a leading entry is 0 except for the leading entry.

So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.

To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?

Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.

Solve:

$x+y=1$

$x+y=2$

Then we have

$\ A = \left( {\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} } \right)$

$\ b = \left( {\begin{array}{cc} 1 \\ 2 \end{array} } \right)$

and $Ax=b$

If we turn A into RREF, we get

$\ E = \left( {\begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} } \right)$

So A has rank 1

and if we put $[A|b]$ into RRE form, we get

$\ E' = \left( {\begin{array}{cc} 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right)$

So augmented matrix has rank 2. Observe what last row says in terms of equations.

• Hmmm, I think I'm starting to understand. Could you provide an example (assuming it wouldn't be too much work to write all the mathjax code for matrices)? Commented Jun 30, 2014 at 22:52
• @user 3002473: It is not too much work, but it means you're going against the family... Commented Jun 30, 2014 at 22:54
• Example of what exactly? If you are asking an example in which rank of $A$ and $[A|b]$ is different, try solving equation x+y=1, x+y=2 using RRE form of matrix. It should not be that difficult to see what is happening Commented Jun 30, 2014 at 22:58
• @user160738 Yeah, an example in which the rank is different between $A$ and $[A|b]$. I'm more of a visual/hands-on learner, so I think I might have a better understanding if I could see the theorems working in a real application. Commented Jun 30, 2014 at 23:01
• Oh! That makes perfect sense now! Thank you so much! Commented Jun 30, 2014 at 23:18