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2.22.2Systems of Linear Equations: Unique SolutionsSystems of Linear Equations: Unique Solutions
3 2 8 9
2 2 1 3
1 2 3 8
3 2 8 9
2 2 1 3
1 2 3 8
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 0 3
0 1 0 4
0 0 1 1
1 0 0 3
0 1 0 4
0 0 1 1
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
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
2
2 2 2
3 1
x y z
y z
y z
2
1
3 1
x y z
y z
y z
2
3 3 0
2 3
x y z
x y z
x y z
Replace R2 with [R1 + R2]
Replace R3 with [–2(R1) + R3]
Replace R2 with ½(R2)
1
2
Row 1 (R1)
Row 2 (R2)
Row 3 (R3)step
. . .
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
2
1
2 4
x y z
y z
z
2
1
2
x y z
y z
z
4
1
2
x y
y
z
4
5
3Replace R3 with [–3(R2) + R3]
Replace R3 with ½(R3)
Replace R2 with [R2 + R3]
Replace R1 with [(–1)R3 + R1]
. . .
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
3
1
2
x
y
z
So the solution is (3, –1, –2).
6
Replace R1 with [R2 + R1]
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.
2
3 3 0
2 3
x y z
x y z
x y z
1 1 1 2
1 3 3 0
2 1 1 3
System Augmented matrix
coefficients constants
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Row Operation Notation
1. Interchange row i and row j
2. Replace row j with c times row j
3. Replace row i with the sum of row i and c times row j
R Ri j
cR j
R cRi j
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Ex. Last example revisited:
2
2 2 2
3 1
x y z
y z
y z
2
3 3 0
2 3
x y z
x y z
x y z
1 1 1 2
1 3 3 0
2 1 1 3
1 1 1 2
0 2 2 2
0 3 1 1
System Matrix
2 1R R
( 2)3 1R R
. . .
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
2
1
2 4
x y z
y z
z
2
1
2
x y z
y z
z
2
1
3 1
x y z
y z
y z
1 1 1 2
0 1 1 1
0 0 2 4
1 1 1 2
0 1 1 1
0 3 1 1
1 1 1 2
0 1 1 1
0 0 1 2
122R
( 3)3 2R R
132R
. . .
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
3
1
2
x
y
z
4
1
2
x y
y
z
1 1 0 4
0 1 0 1
0 0 1 2
1 0 0 3
0 1 0 1
0 0 1 2
( 1)1 3R R
2 3R R
1 2R R
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
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 5 1
0 1 0 3
0 0 1 5
1 0 0 8
0 1 0 4
0 0 0 0
Row-Reduced Form Non Row-Reduced Form
R2 , R3 switched
Must be 0
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
2 123 1
1 1 1 2 1 1 1 2
1 3 3 0 0 2 2 2
2 1 1 3 0 3 1 1
R RR R
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.
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
1 0 9 12
0 1 19 / 2 27 / 2
0 0 31 31
1 0 9 12
0 1 19 / 2 27 / 2
0 0 31 31
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 2 / 3 8 / 3 3
0 2 / 3 19 / 3 9
0 8 / 3 17 / 3 5
1 2 / 3 8 / 3 3
0 2 / 3 19 / 3 9
0 8 / 3 17 / 3 5
Example 5, page 82-83
1 0 0 3
0 1 0 4
0 0 1 1
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.
3 2 8 9
2 2 1 3
1 2 3 8
3 2 8 9
2 2 1 3
1 2 3 8
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
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:
Use the Gauss-Jordan elimination method,
we have . . . . . .
200
40 60 80 12,800
20 25 40 6100
x y z
x y z
x y z
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Solution (cont.)
. . .. . .
1 1 1 200 1 0 0 50
40 60 80 12800 0 1 0 60
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.
