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.




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.

  • $\begingroup$ 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)? $\endgroup$ Jun 30 '14 at 22:52
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    $\begingroup$ @user 3002473: It is not too much work, but it means you're going against the family... $\endgroup$
    – user99680
    Jun 30 '14 at 22:54
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    $\begingroup$ 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 $\endgroup$
    – user160738
    Jun 30 '14 at 22:58
  • $\begingroup$ @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. $\endgroup$ Jun 30 '14 at 23:01
  • $\begingroup$ Oh! That makes perfect sense now! Thank you so much! $\endgroup$ Jun 30 '14 at 23:18

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