Linear Programming Lecture Notes - Penn State Personal Web Server

Definition 3.46 (Basic Variables). For historical reasons, the variables in the vector
xB are called the basic variables and the variables in the vector xN are called the non-basic
variables.
We can use matrix multiplication to expand the left hand side of this expression as:
(3.50) BxB + NxN = b
The fact that B is composed of all linearly independent columns implies that applying Gauss-
Jordan elimination to it will yield an m × m identity and thus that B is invertible. We can
solve for basic variables xB in terms of the non-basic variables:
(3.51) xB = B −1 b − B −1 NxN
We can find an arbitrary solution to the system of linear equations by choosing values for
the variables the non-basic variables and solving for the basic variable values using Equation
3.51.
Definition 3.47. (Basic Solution) When we assign xN = 0, the resulting solution for x
is called a basic solution and
(3.52) xB = B −1 b
Example 3.48. Consider the problem:
1 2 3
(3.53)
4 5 6
⎡
⎣ x1
⎤
x2⎦
7
=
8
x3
Then we can let x3 = 0 and:
1 2
(3.54) B =
4 5
We then solve1 :
x1
(3.55) = B −1
x2
7
=
8
−19
3
20
3
Other basic solutions could be formed by creating B out of columns 1 and 3 or columns
2 and 3.
and
Exercise 38. Find the two other basic solutions in Example 3.48 corresponding to
2 3
B =
5 6
B =
1 3
4 6
In each case, determine what the matrix N is. [Hint: Find the solutions any way you like.
Make sure you record exactly which xi (i ∈ {1, 2, 3}) is equal to zero in each case.]
1 Thanks to Doug Mercer, who found a typo below that was fixed.
46