16. Echelon examples
Take the system The augmented matrix is To clear the first column, we do and , which gives: Now we do and to get: Finally, clear column 2 with and , giving: This is now in reduced echelon form. It corresponds to the equations The third equation is trivially satisfied: it is just a check that our equation really has a solution. The other two can be used to express the dependent variables in terms of the free variable . Note that are chosen as dependent variables because they are the variables whose coefficients are leading entries.
Take the system The augmented matrix is We do , yielding: Now we clear column 1 with and , which gives: Finally, yields: The left-hand block is now in reduced echelon form.
Translating this back into equations gives This means there is no solution, because .
Take the very similar system Here, is a parameter; we'll perform the same analysis and figure out what value needs to take in order for the system to have a solution. The augmented matrix is The same sequence of row operations yield We see this only has a solution if . In this case, the solution is , ( is free).
The moral of this is: for an augmented matrix with in reduced echelon form,
any row of zeros in gives a constraint on the vector which has to be satisfied if a solution is to exist,
any other row gives you a way of expressing a single dependent variable in terms of free variables.