16. Echelon examples

Example 1

0.00 Take the system x + 2 y + z = 5, minus x + y + 2 z = 1, x minus z = 1. The augmented matrix is 1, 2, 1, bar, 5; minus 1, 1, 2, bar, 1; 1, 0, minus 1, bar, 1. To clear the first column, we do R_2 maps to R_2 + R_1 and R_3 maps to R_3 minus R_1, which gives: 1, 2, 1, bar, 5; 0, 3, 3, bar, 6; 0, minus 2, minus 2, bar, minus 4. 2.50 Now we do R_2 maps to R_2 over 3 and R_3 maps to minus a half R_3 to get: 1, 2, 1, bar, 5; 0, 1, 1, bar, 2; 0, 1, 1, bar, 2. Finally, clear column 2 with R_3 maps to R_3 minus R_2 and R_1 maps to R_1 minus 2 R_2, giving: 1, 0, minus 1, bar, 1; 0, 1, 1, bar, 2; 0, 0, 0, bar, 0. 5.33 This is now in reduced echelon form. It corresponds to the equations x minus 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

6.50 Take the system 2 x + 2 y = 2, x + y + z = 8, x + y minus z = minus 5. The augmented matrix is 2, 2, 0, bar, 2; 1, 1, 1, bar, 8; 1, 1, minus 1, bar, minus 5. We do R_1 maps to a half R_1, yielding: 1, 1, 0, bar, 1; 1, 1, 1, bar, 8; 1, 1, minus 1, bar, minus 5. 8.04 Now we clear column 1 with R_2 maps to R_2 minus R_1 and R_3 maps to R_3 minus R_1, which gives: 1, 1, 0, bar, 1; 0, 0, 1, bar, 7; 0, 0, minus 1, bar, minus 6. 9.06 Finally, R_3 maps to R_3 + R_2 yields: 1, 1, 0, bar, 1; 0, 0, 1, bar, 7; 0, 0, 0, bar, 1. The left-hand block is now in reduced echelon form.

9.44 Translating this back into equations gives x + y = 1 , z = 7 and 0 = 1. This means there is no solution, because 0 is not equal to 1.

Example 3

11.07 Take the very similar system 2 x + 2 y = 2, x + y + z = 8, x + y minus 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, bar, 2; 1, 1, 1, bar, 8; 1, 1, minus 1, b. The same sequence of row operations yield 1, 1, 0, bar, 1; 0, 0, 1, bar 7; 0, 0, 0, bar, b + 6. 13.11 We see this only has a solution if b equals minus 6. In this case, the solution is x = 1 minus y, z = 7 (y is free).

13.57 The moral of this is: for an augmented matrix M bar 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.