Page 48 - A Course in Linear Algebra with Applications

32 Chapter Two: Systems of Linear Equations

At this point we should have liked X2 to appear in the second
equation: however this is not the case. To remedy the situ-
ation we interchange equations 2 and 3, in symbols (2)«->(3).
The linear system now takes the form

xi + 4x 2 + 2x 3 = - 2
X 2 + X 3 = 1
7x 3 = 28

|
|
Finally, multiply equation 3 by , that is, apply (3), to get
xi + 4x 2 + 2x 3 = - 2
+ = 1
x 2 x 3
x 3 = 4
This system can be solved quickly by a process called back
substitution. By the last equation X3 = 4, so we can substi-
tute X3 = 4 in the second equation to get x 2 = —3. Finally,
substitute X3 = 4 and x 2 = — 3 in the first equation to get
x\ = 2. Hence the linear system has a unique solution.

Example 2.1.3

( Xi + 3x 2 + 3x 3 + 2x 4 = 1
< 2X! + 6x 2 + 9x 3 + 5X4 = 5
1 -xi - 3x 2 + 3x 3 = 5

Apply operations (2) - 2(1) and (3) + (1) successively to
the linear system to get

Xi + 3x 2 3x 3 + 2x 4 = 1
3x 3 + x 4 = 3
= 6
6x3 + 2x 4

Since X2 has disappeared completely from the second and third
|
equations, we move on to the next unknown x 3 ; applying (2),
we obtain