Linear Algebra, Theory And Applications, 2012a

30 PRELIMINARIES
Definition 1.10.5 Since z = t where t is arbitrary, the variable z is called a free variable.
The phenomenon of an infinite solution set occurs in equations having only one variable
also. For example, consider the equation x = x. It doesn’t matter what x equals.
Definition 1.10.6 A system of linear equations is a list of equations,
n∑
a ij x j = f j , i =1, 2, 3, ··· ,m
j=1
where a ij are numbers, f j is a number, and it is desired to find (x 1 , ··· ,x n ) solving each of
the equations listed.
As illustrated above, such a system of linear equations may have a unique solution, no
solution, or infinitely many solutions. It turns out these are the only three cases which can
occur for linear systems. Furthermore, you do exactly the same things to solve any linear
system. You write the augmented matrix and do row operations until you get a simpler
system in which it is possible to see the solution. All is based on the observation that the
row operations do not change the solution set. You can have more equations than variables,
fewer equations than variables, etc. It doesn’t matter. You always set up the augmented
matrix and go to work on it. These things are all the same.
Example 1.10.7 Give the complete solution to the system of equations, −41x +15y = 168,
109x − 40y = −447, −3x + y =12, and 2x + z = −1.
The augmented matrix is
⎛
⎜
⎝
−41 15 0 168
109 −40 0 −447
−3 1 0 12
2 0 1 −1
To solve this multiply the top row by 109, the second row by 41, add the top row to the
second row, and multiply the top row by 1/109. Note how this process combined several
row operations. This yields
⎛
⎜
⎝
−41 15 0 168
0 −5 0 −15
−3 1 0 12
2 0 1 −1
Next take 2 times the third row and replace the fourth row by this added to 3 times the
fourth row. Then take (−41) times the third row and replace the first row by this added to
3 times the first row. Then switch the third and the first rows. This yields
⎛
⎜
⎝
123 −41 0 −492
0 −5 0 −15
0 4 0 12
0 2 3 21
Take −1/2 times the third row and add to the bottom row. Then take 5 times the third
row and add to four times the second. Finally take 41 times the third row and add to 4
times the top row. This yields
⎛
⎜
⎝
492 0 0 −1476
0 0 0 0
0 4 0 12
0 0 3 15
⎞
⎟
⎠ .
⎞
⎟
⎠ .
⎞
⎟
⎠ .
⎞
⎟
⎠