16. Echelon examples

16. Echelon examples

Example 1

Take the system x + 2 y + z = 5 , - x + y + 2 z = 1 , x - z = 1 . The augmented matrix is ( 1 2 1 | 5 - 1 1 2 | 1 1 0 - 1 | 1 ) . To clear the first column, we do R 2 R 2 + R 1 and R 3 R 3 - R 1 , which gives: ( 1 2 1 | 5 0 3 3 | 6 0 - 2 - 2 | - 4 ) . Now we do R 2 1 3 R 2 and R 3 - 1 2 R 3 to get: ( 1 2 1 | 5 0 1 1 | 2 0 1 1 | 2 ) . Finally, clear column 2 with R 3 R 3 - R 2 and R 1 R 1 - 2 R 2 , giving: ( 1 0 - 1 | 1 0 1 1 | 2 0 0 0 | 0 ) . This is now in reduced echelon form. It corresponds to the equations x - z = 1 , y + z = 2 , 0 = 0 . 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 x , y in terms of the free variable z . Note that x , y are chosen as dependent variables because they are the variables whose coefficients are leading entries.

Example 2

Take the system 2 x + 2 y = 2 , x + y + z = 8 , x + y - z = - 5 . The augmented matrix is ( 2 2 0 | 2 1 1 1 | 8 1 1 - 1 | - 5 ) . We do R 1 1 2 R 1 , yielding: ( 1 1 0 | 1 1 1 1 | 8 1 1 - 1 | - 5 ) . Now we clear column 1 with R 2 R 2 - R 1 and R 3 R 3 - R 1 , which gives: ( 1 1 0 | 1 0 0 1 | 7 0 0 - 1 | - 6 ) . Finally, R 3 R 3 + R 2 yields: ( 1 1 0 | 1 0 0 1 | 7 0 0 0 | 1 ) . The left-hand block is now in reduced echelon form.

Translating this back into equations gives x + y = 1 , z = 7 , 0 = 1 . This means there is no solution, because 0 1 .

Example 3

Take the very similar system 2 x + 2 y = 2 , x + y + z = 8 , x + y - z = b . Here, b is a parameter; we'll perform the same analysis and figure out what value b needs to take in order for the system to have a solution. The augmented matrix is ( 2 2 0 | 2 1 1 1 | 8 1 1 - 1 | b ) . The same sequence of row operations yield ( 1 1 0 | 1 0 0 1 | 7 0 0 0 | b + 6 ) . We see this only has a solution if b = - 6 . In this case, the solution is x = 1 - y , z = 7 ( y is free).

The moral of this is: for an augmented matrix ( M | v ) with M in reduced echelon form,

  • any row of zeros in M gives a constraint on the vector v 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.