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2.2
Systems of Linear Equations: Unique Solutions
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
 3 2 8 9 



2
2
1
3


 1 2 3 8 
1

0
 0
0
1
0
0
0 1
3

4
1 
The Gauss-Jordan Elimination Method Operations
1. Interchange any two equations.
2. Replace an equation by a nonzero constant multiple
of itself.
3. Replace an equation by the sum of that equation and a
constant multiple of any other equation.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Solve the system
2x  y  z  3
step
1
x yz 2
2y  2 z  2
3y  z  1
2
x y z  2
 x  3 y  3z  0
x yz 2
y  z 1
Row 1 (R1)
Row 2 (R2)
Row 3 (R3)
Replace R2 with [R1 + R2]
Replace R3 with [–2(R1) + R3]
Replace R2 with ½(R2)
3 y  z  1
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
x yz  2
3
y  z 1
2 z  4
Replace R3 with [–3(R2) + R3]
x yz 2
4
y  z 1
z  2
x y
5
y
Replace R3 with ½(R3)
4
Replace R1 with [(–1)R3 + R1]
 1
Replace R2 with [R2 + R3]
z  2
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
3
x
6
y
Replace R1 with [R2 + R1]
 1
z  2
So the solution is (3, –1, –2).
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Augmented Matrix
*Notice that the variables in the preceding example
merely keep the coefficients in line. This can also be
accomplished using a matrix. A matrix is a rectangular
array of numbers.
System
x y z  2
 x  3 y  3z  0
2x  y  z  3
Augmented matrix
 1 1 1 2 



1
3

3
0


 2 1 1 3 
coefficients
constants
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Row Operation Notation
1. Interchange row i and row j
Ri  R j
2. Replace row j with c times row j
cR j
3. Replace row i with the sum of row i and c times row j
Ri  cR j
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Last example revisited:
Matrix
System
x y z  2
 x  3 y  3z  0
 1 1 1 2 



1
3

3
0


 2 1 1 3 
2x  y  z  3
x yz  2
2y  2 z  2
3y  z  1
R2  R1
R3  ( 2) R1
1 1 1 2 


0
2

2
2


0 3 1 1
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
x yz 2
y  z 1
1
R
2 2
3 y  z  1
x yz  2
y  z 1
2 z  4
R3  ( 3) R2
x yz 2
y  z 1
z  2
1
R
2 3
 1 1 1 2 


0 1 1 1 
0 3 1 1
1 1 1 2 


0 1 1 1 
0 0 2 4 
1 1 1 2 


0
1

1
1


0 0 1 2 
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
x y
y
4
 1
R1  ( 1) R3
R2  R3
z  2
3
x
y
 1
R1  R2
z  2
1 1 0 4 


0 1 0 1
0 0 1 2 
1 0 0 3 


0
1
0

1


0 0 1 2 
This is in Row- Reduced
Form
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Row–Reduced Form of a Matrix
1. Each row consisting entirely of zeros lies below any
other row with nonzero entries.
2. The first nonzero entry in each row is a 1.
3. In any two consecutive (nonzero) rows, the leading 1
in the lower row is to the right of the leading 1 in the
upper row.
4. If a column contains a leading 1, then the other entries
in that column are zeros.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Row–Reduced Form of a Matrix
Row-Reduced Form
Non Row-Reduced Form
1 0 0 3 


0 1 0 1
0 0 1 2 
1 0 0 9 


0 0 1 4 
0 1 0 2 
1 0 0 8 


0 1 0 4 
0 0 0 0 
R2 , R3
switched
Must
be 0
1 0 5 1 


0 1 0 3 
0 0 1 5
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Unit Column
A column in a coefficient matrix where one of the entries is 1 and
the other entries are 0.
1 0 5 1 


0
1
0
3


0 0 1 5
Unit columns
Not a Unit
column
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Pivoting – Using a coefficient to transform a
column into a unit column
 1 1 1 2 
1 1 1 2 

 R2  R1 


1
3

3
0


0
2

2
2

 R3 2 R1 

 2 1 1 3 
0 3 1 1
This is called pivoting on the 1 and it is
circled to signify it is the pivot.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Gauss-Jordan Elimination Method
1. Write the augmented matrix
2. Interchange rows, if necessary, to obtain a nonzero first
entry. Pivot on this entry.
3. Interchange rows, if necessary, to obtain a nonzero
second entry in the second row. Pivot on this entry.
4. Continue until in row-reduced form.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Use the Gauss-Jordan elimination method to solve the system
of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
 3 2 8

1
 2 2
 1 2 3
9

3
8 
 1 2 / 3
8/3

0 2 / 3 19 / 3
0 8 / 3 17 / 3
3

9
5
1 0
9
12 


0
1
19
/
2
27
/
2


0 0 31 31 
 1 0 0 3


0
1
0
4


 0 0 1 1 
The solution to the system is thus x = 3, y = 4, and z = 1.
Example 5, page 82-83
Example
A farmer has 200 acres of land suitable for cultivating crops A, B,
and C. The cost per acre of cultivating crop A, crop B, and crop C
is $40, $60, and $80, respectively. The farmer has $12,800
available for land cultivation. Each acre of crop A requires 20
labor-hours, each acre of crop B requires 25 labor-hours, and each
acre of crop C requires 40 labor-hours. The farmer has a
maximum of 6100 labor-hours available. If he wishes to use all of
his cultivatable land, the entire budget, and all of labor available,
how many acres of each crop should he plant?
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Solution
Let x = the number of acres of crop A
Let y = the number of acres of crop B
Let z = the number of cares of crop C
Then we have:
x 
y  z 
200
40 x  60 y  80 z  12,800
20 x  25 y  40 z  6100
Use the Gauss-Jordan elimination method,
we have
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Solution (cont.)
 1 1 1 200 
1 0 0 50 




 0 1 0 60 
 40 60 80 12800  
 20 25 40 6100 
 0 0 1 90 
From the last augmented matrix in reduced form, we see that
x = 50, y = 60, and z = 90. Therefore, the farmer should
plant 50 acres of crop A, 60 acres of crop B, and 90 acres
of crop C.
...
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.