A matrix is in reduced echelon form if:
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it's in echelon form,
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all its leading entries are equal to 1 ,
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(*) in any column containing a leading entry, there are no other nonzero entries.
A matrix is in reduced echelon form if:
it's in echelon form,
all its leading entries are equal to 1 ,
(*) in any column containing a leading entry, there are no other nonzero entries.
Consider the following matrices: A=(123012001),B=(11),C=(1011012-1)
A is in echelon form an its leading entries are all 1 . However, condition (*) doesn't hold: for example column 2 contains a leading entry (1 ) and another nonzero entry (the 2 above it).
B is in echelon form, its leading entry is 1 and, in the column containing the leading entry, there are no other nonzero entries, so B is in reduced echelon form.
C is in reduced echelon form.
D and F are not even in echelon form.
E is in echelon form, but its leading entries are not 1 , so it is not in reduced echelon form.
Consider the augmented matrix (1011|5012-1|13).
We get w=5-y-z and x=13-2y+z , using w,x as dependent variables and y,z as free variables. Why did I pick w,x as the dependent variables? If we had tried to use the first equation to express y in terms of w and z , since y appears in the second equation, we would need to substitute y=5-w-z into the second equation to proceed. We didn't need to do that for w or x because w only appeared in equation 1 and x only appeared in equation 2. That is a consequence of condition (*): if we take our dependent variables to be the ones whose coefficients are the leading entries then they only appear in one of the equations.
Consider the augmented matrix (1201|10018|20000|0) . This corresponds to the equations: w+2x+z=1,y+8z=2,0=0.
Note that x also appears only in the first equation, so we could use x as a dependent variable, but that's just luck: we're only guaranteed that the variables whose coefficients are leading entries will appear in precisely one equation.
Consider the augmented matrix (1201|10018|20000|3) which differs from the previous one only in the very last entry. That corresponds to the system: w+2x+z=1,y+8z=2,0=3.
Our strategy (which we'll see implemented in some examples next time) is as follows:
given a system of equations, write down an augmented matrix;
perform row operations on the augmented matrix until the part left of the bar is in reduced echelon form;
read off from this whether solutions exist (they do unless one of the rows is (0⋯0|b) for b≠0 );
if solutions exist, write down the solution by finding the dependent variables in terms of the free variables.