Linear Programming Lecture Notes - Penn State Personal Web Server

and right hand side vector b = [7, 8, 9] T . Applying Gauss-Jordan elimination in this case
yields:
⎡
⎤
19
1 0 −1 − 3
(3.28) X := ⎣ ⎦
0 1 2 20
3
0 0 0 0
Since the third row is all zeros, there are an infinite number of solutions. An easy way to
solve for this set of equations is to let x3 = t, where t may take on any value in R. Then,
row 2 of Expression 3.28 tells us that:
(3.29) x2 + 2x3 = 20
3 =⇒ x2 + 2t = 20
3 =⇒ x2 = 20
− 2t
3
We then solve for x1 in terms of t. From row 1 of Expression 3.28 we have:
(3.30) x1 − x3 = − 19
3 =⇒ x1 − t = − 19
3 =⇒ x1 = t − 19
3
Thus every vector in the set:
(3.31) X = t − 19
20
, − 2t, t
3 3
: t ∈ R
is a solution to Ax = b.
Conversely, suppose we have the problem:
x1 + 2x2 + 3x3 = 7
4x1 + 5x2 + 6x3 = 8
7x1 + 8x2 + 9x3 = 10
The new right hand side vector is b = [7, 8, 20] T . Applying Gauss-Jordan elimination in this
case yields:
(3.32)
⎡
1
X := ⎣ 0
0
1
−1
2
⎤
0
0 ⎦
0 0 0 1
Since row 3 of X has a non-zero element in the b ′ column, we know this problem has no
solution, since there is no way that we can find values for x1, x2 and x3 satisfying:
(3.33) 0x1 + 0x2 + 0x3 = 1
Exercise 30. Solve the problem
x1 + 2x2 = 7
3x1 + 4x2 = 8
using Gauss-Jordan elimination.
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