Finite Mathematics
Test #1
March 21, 2001
 
R. Hammack
Name: ________________________  
Score: _________

(1) Use Gauss - Jordan elimination to solve the following system of equations.

x1
+ 2x2
- 3x3
+ x4
=
-2
3x1
- x2
- 2x3
- 4x4
=
1
2x1
+ 3x2
- 5x3
+ x4
=
-3

First we set up the augmented matrix.
[
1
2
-3
1
|
|
|
|
-2
 ]
3
-1
-2
-4
1
2
3
-5
1
-3

-3R1 + R2 ---> R2
2R1 + R3 --> R3
[
1
2
-3
1
|
|
|
|
-2
 ]
0
-7
7
-7
7
0
-1
1
-1
1

-1/7 R2 ---> R2
[
1
2
-3
1
|
|
|
|
-2
 ]
0
1
-1
1
-1
0
-1
1
-1
1

-2R2 + R1 ---> R1
R2 + R3 ---> R3
[
1
0
-1
-1
|
|
|
|
0
 ]
0
1
-1
1
-1
0
0
0
0
0

This matrix corresponds to the system ...
x1
- x3
- x4
=
0
x2
- x3
+ x4
=
-1
     
0
=
0

.. which is the same as ...
x1
= x3 + x4
x2
= -1 + x3 - x4

Therefore the solutions are:
x1
= s + t
x2
= -1 + s - t
x3
 = s
x4
= t

(2)

(a) Multiply the matrices:

[
2
-1
][
3
1
1
 ] =
4
1
2
4
0
= [
(2)(3)-(1)(2)
(2)(1)-(1)(4)
(2)(1)-(1)(0)
 ]
(4)(3) +(1)(2)
(4)(1) +(1)(4)
(4)(1) +(1)(0)
= [
4
-2
2
 ]
14
8
4

(b) Find values for x and y that make the following expression true.

[
1
2x
] + [
4
3y
] = [
5
-3
]
5x
3
-y
2
18
5

Adding, we get
[
5
2x +3y
] = [
5
-3
 ]
5x - y
5
18
5

For these two matrices to be equal, we need to have 2x + 3y = -3 and 5x -y = 18.
Thus to answer the question, we need to solve the following system.

2x + 3y = -3
5x -y = 18.

Substitution should work here. The second equation gives y = 5x - 18, and plugging this in for y in the first equation gives
2x + 3(5x - 18) = -3
2x + 15x - 54 = -3
17x = 51
x = 51/17 = 3

Thus x = 3, and to find y we plug this value into the equation y = 5x - 18.
y = 3(3) - 18 = -3.

Thus the answer is x = 3, y = -3.

(3) Graph the solutions of the following system of inequalities. Be sure to find and label all the corner points.

2x1 - 3x2
≥ -6
4x1 + 3x2
≥ 24
x1
≥ 0
x2
≥ 0

The system is graphed on the right.
The feasible region is unbounded.
There are just two corner points (6, 0) and (3, 4).

(4) Solve the following linear programming problem by graphing. Do not use the simplex method. (Note: You sketched the feasible region in the previous problem. Feel free to use that informatation here.)

Maximize
P = 2x1 + x2
Subject to ... 2x1 - 3x2 ≥ -6
4x1 + 3x2 ≥ 24
x1 ≥ 0
x2 ≥ 0

The minimum will occur at a corner point, and from the work above there are just two corner points.

Corner points
P = 2x1 + x2
(3, 4)
P = (2)(3) + 4 = 10
(6, 0)
P = (2)(6) + 0 = 12

The minimum value of P = 10 thus occurs at the corner point (3, 4). Thus P is minimized at 10 when x1 = 3 and x2 = 4.

(5) Use the simplex method to solve the following linear programming problem.

Maximize
P = 6x1 + 4x2 + x3
Subject to ...
2x1 + 4x2 + x3 ≤ 26
x1 , x2 , x3 ≥ 0

Introducing slack variables, this becoms:

2x1
+ x2
+ 2x3
+ s1
= 14
2x1
+ 4x2
+ x3
+ s2
= 26
- 6x1
- 4x2
- x3
+ P
 = 0
Now we set up the simplex tableau.
The pivot rows and columns are indicated.
The pivot operation begins.
   
x1
x2
x3
s1
s2
P
   
s1  
2
1
2
1
0
0
|
14
   14 / 2 = 7
s2
2
4
1
0
1
0
|
26
 26 / 2 = 13
 
 
P
-6
-4
-1
0
0
1
|
0
  

1/2 R1 ---> R1
   
x1
x2
x3
s1
s2
P
   
   
1
1/2
1
1/2
0
0
|
7
  
 
2
4
1
0
1
0
|
26
  
 
 
 
-6
-4
-1
0
0
1
|
0
  

-2R1 + R2 ---> R2
6R1 + R3 --> R3
   
x1
x2
x3
s1
s2
P
   
x1  
1
1/2
1
1/2
0
0
|
7
   7 / 1/2 = 14
s2
0
3
-1
-1
1
0
|
12
 12 / 3 = 4
 
 
P
0
-1
5
3
0
1
|
42
  

The first pivot operation is complete, and there is still a negative indicator. We thus do another pivot operation. The pivot column and row are picked (above), and the second pivot operation begins.
1/3 R2 ---> R2
   
x1
x2
x3
s1
s2
P
   
   
1
1/2
1
1/2
0
0
|
7
    
 
0
1
-1/3
-1/3
1/3
0
|
4
  
 
 
 
0
-1
5
3
0
1
|
42
  

-1/2 R2 + R1 ---> R1
R2 + R3 ---> R3
   
x1
x2
x3
s1
s2
P
   
x1  
1
0
7/6
2/3
-1/6
0
|
5
    
x2
0
1
-1/3
-1/3
1/3
0
|
4
  
 
 
P
0
0
14/3
7/6
8/3
1
|
46
  
Now there are no more negative indicators, so we are done. The optimal solution can be read off the table.
P is maximized at 46 when x1 = 5, x2 = 4, and x3 = 0.