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 $\left[\begin{array}{c|c}A&b\end{array}\right]$?
Best Answer
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.