Section 2.3 Gauss-Jordan Method
Section 2.3 Gauss-Jordan Method
Section 2.3
1.In reduced echelon form
2.Not in reduced echelon form because the leading non-zero number in row 3 is not 1.
3.Not in reduced echelon form because column 3 does not contain a zero in row 2.
4.In reduced echelon form.5.In reduced echelon form.
6.Not in reduced echelon form. The row of zeros should be the bottom row.
7.Not in reduced echelon form because the leading 1 in row 3 is to the left of the leading 1 in row 2, and the 3 in row 3 should be 0.
8.In reduced echelon form.
9.Perform the following row operations to reduce the third column.
R3 R3, -2R3 + R1 R1, and -3R3 + R2 R2 and obtain
10.The next step is to get a 1 in the (2, 2) position and zeros in the rest of column 2. This is done with the sequence of row operations.
R2 R2, 3R2 + R1 R1, and 5R2 + R3 R3 which gives
11.Exchange rows 2 and 3, then get a zero in the (1, 2) position with R2 R3 and -2R2 + R1 R1 giving
12.Get a 1 in the (1, 1) position and zero the rest of column 1 with the row operations R1 R2 and R1 giving then -3R1 + R3 R3 giving
13.
Get a 1 in the (2, 2) position and 0 in (1, 2) position with R2 R3 and -3R2 + R1 R1 giving
14.Interchange rows 3 and 4 and get zeros in column 4, rows 1 and 2 with R3 R4, -2R3 + R1 R1, and - R3 + R2 R2 giving
15.
16.
17.
18.
19.
20.
21.x1 = 322.x1 + 3x3 = -1
x2 = -2x2 + 2x3 = 5
x3 = 5There are an infinite number of solutions
The solution is (3, -2, 5). of the form (-1 - 3k, 5 - 2k, k).
23.x1 + 3x3 = 424.x1 + 2x3 + 5x5 = 3
x2 + x3 = -6x2 - x3 + 4x5 = 9
x4 = 2x4 + 3x5 = 7
An infinite number of solutions A infinite number of solutions of the form
of the form (4 - 3k, -6 - k, k, 2). (3 - 2k - 5m, 9 + k - 4m, k, 7 - 3m, m).
25.x1 = 026.x1 + 2x4 = 0
x2 = 0x2 - 3x4 = 0
0 = 1x3 + 5x4 = 0
No solution 0 = 1
No solution
27.x1 + 3x4 = 028.x1 + 2x3 + 3x4 = 0
x2 - 2x4 = 0x2 - x3 + 5x4 = 0
x3 + 7x4 = 00 = 0
0 = 00 = 0
There are an infinite number of solutionsThere are an infinite number of solutions
of the form (-3k, 2k, -7k, k)of the form (-2k - 3m, k - 5m, k, m).
29.No solution
30.
x1 = 24 + x3, x2 = -13 - x3 or (24 + k, -13 - k, k)
31.
x1 = 4, x2 = -1/2, x3 = -3/2
32.
x1 = 2, x2 = -1 - 2x3, x4 = 3 or (2, -1 - 2k, k, 3)
33.
x1 = -32 - 2x3, x2 = 45 + x3, x4 = 16 or (-32 - 2k, 45 + k, k, 16)
34.
x1 = -6x2 + x4, x3 = -3x4 or (-6k + r, k, -3r, r)
35.No solution
36.
No solution
37.No solution
38.No solution
39.
x1 = -59 + 11x3, x2 = 23 - 5x3 or (-59 + 11k, 23 - 5k, k)
40.
x1 = -1 - x3, x2 = 1 + 2x3 or (-1 - k, 1 + 2k, k)
41.
x1 = 3 + x3 + x4, x2 = 6 + x3 + x4 - x5 or (3 + k + r, 6 + k + r - s, k, r, s)
42.
x1 = 3 + 38x3, x2 = -1 + 14x3 or (3 + 38k, -1 + 14k, k)
43.No solution
44. No solution
45.No solution
46.No solution
47.x = 2, y = -3
48.No solution
49.x = -5, y = 2
50.x = 2, y = 2
51. x1 = 2, x2 = 1, x3 = 3
52.No solution
53.
No solution
54.
x1 = 5 + x3, x2 = 1 - x3 or (5 + k, 1 - k, k)
55.No solution
56. (2, 3, -1)
57.
x1 = , x2 = -, x3 = 0, x4 =
58.No solution
59.
(-9, 6, 2, 1)
60.
x1 = -2x2 + x5, x3 = -3x5, x4 = -2x5 or (-2k + r, k, -3r, -2r, r)
61.
x1 = -2x2 + 3x3 - x5 + 3x6, x4 = -2x5 - x6 or (-2k + 3r - s + 3t, k, r, -2s - t, s, t)
62.Let x1 = number of portfolio I, x2 = number of portfolio II, x3 = number of portfolio III.
3x1 + x2 + 5x3 = 50
2x1 + 4x2 + 10x3 = 160
x1 + x2 + 3x3 = 25
No solution
63.Let x1 = amount in stocks, x2 = amount in bonds, x3 = amount in money markets.
x1 + x2 + x3 = 45,000
-2x1 + x2 + x3 = 0
0.10x1 + 0.07x2 + 0.075x3 = 3,660
$15,000 in stocks, $18,000 in bonds, $12,000 in money market
64.Let x = number of houses, y = number of duplexes, z = number of apartments.
x + y + z = 250
4500x + 4000y + 3000z = 875000
10x + 12y + 6z = 2050
40 houses, 65 duplexes, 145 apartments
65.Let x = minutes jogging, y = minutes playing handball, z = minutes biking.
x + y + z = 60
13x + 11y + 7z = 660
x - 2z = 0
x = 2z, y = 60 - 3z
She may bike from 0 to 20 min., then should jog for twice that time, and play handball for the rest of the 60 minutes.
66.Let x = number of roosters, y = number of hens, z = number of chicks.
x + y + z = 100
5x + 3y + 0.05z = 100
x = 1.475z - 100, y = 200 - 2.475z
Now x ³ 1 and y ³ 1, so
1.475z - 100 ³ 1 and 200 - 2.475z ³ 1
z ³ 68.47 and z ² 80.40
Since the total cost is an even number of dollars and chicks cost $0.05 each, the number of chicks must be a multiple of 20. The only multiple of 20 that satisfies 68.47 ² z ² 80.4 is z = 80.
18 roosters, 2 hens, and 80 chicks
67.Let x1 = hours of Math, x2 = hours of English, x3 = hours of Chemistry, x4 = hours of History
x1 + x2 + x3 + x4 = 42
x1 + x2 = 21
x1 = 2x2
x2 = 2x4
14 hours for Math, 7 hours for English, and 17.5 hours for Chemistry, and 3.5 hours for History.
68.Let x1 = number of Big Burgers, x2 = number of French Fries, and x3 = number of soft drinks
710x1 + 360x2 + 230x3 = 9360(Calories)
451 + 18x2 = 477(Grams of fat)
9x1 + x2 + 56x3 = 352(Grams of sugar)
The augmented matrix of the system is:
The reduced echelon form is:
They ordered 7 Big Burgers, 9 orders of French Fries, and 5 soft drinks.
69.Let x1 = federal tax, x2 = state tax, and x3 = city tax.
x1 = 0.40(58,400 - x2 - x3)
x2 = 0.20(58,400 - x1 - x3)
x3 = 0.10(58,400 - x1 - x2)
We can reduce these equations to:
x1 + 0.40x2 + 0.40x3 = 23,360
0.20x1 + x2 + 0.20x3 = 11,680
0.10x1 + 0.10x2 + x3 = 5,840
The reduced echelon form of the augmented matrix of this system is
The federal tax is $19,200, the state tax is $7200, and the city tax is $3200.
70. (a)Let x1 = cost of a CD, x2 = cost of a DVD movie, and x3 = cost of a video cassette.
4x1 + x2 + 2x3 = 47(Joshua’s purchase)
3x1 + 2x2 + x3 = 45(Shilpa’s purchase)
The reduced echelon form of the augmented matrix is
so the system has an infinite number of solutions.
(b)The system of equations is
4x1 + x2 + 2x3 = 47(Joshua’s purchase)
3x1 + 2x2 + x3 = 45(Shilpa’s purchase)
x1 + 3x2 + x3 = 46(Lilia’s purchase)
The reduced echelon form of the augmented matrix is
A CD cost $5, a DVD movie cost $11, and a video cassette cost $8.
71.Let x1 = cost of a CD, x2 = cost of a DVD movie, and x3 = cost of a cassette tape.
6x1 + 2x2 + 4x3 = 40(Amy’s purchase)
3x1 + 6x2 + x3 = 53(Bill’s purchase)
6x1 + 7x2 + 3x3 = 73(Carlton’s purchase)
The reduced echelon form of the augmented matrix is
This indicates an infinite number of solutions where
x1 = 4.4666 - 0.7333x3 which indicates x1 4.466
x2 = 6.6 + 0.2x3 which indicates x2 ³ 6.6
Since x1 must be greater than or equal to zero, 4.4666 - 0.7333x3 ³ 0 or x36.09
72.Let x = number of fiction, y = number of nonfiction, and z = number of reference books.
x + y + z = 500
x = z + 50
30x + 40y + 50z = 19500
(a)The augmented matrix of the system of equations is
The reduced form is
Thus the system has many solutions
x = z + 50
y = –2z + 450
(b)Since x, y, z are not negative,
–2z + 450 ³ 0
450 ³ 2z
225 ³ z
x = z + 50 implies x ³ 50
z has a minimum value of 0 and a maximum of 225.
x has a minimum value of 50 when z = 0 and a maximum value of 275 when z = 225.
y has a minimum value of 0 when z = 225 and a maximum value of 450 when z = 0.
73.(a)Let x1 = number supplied by Sweats-Plus to Spirit Shop 1
x2 = number supplied by Sweats-Plus to Spirit Shop 2
x3 = number supplied by Sweats-Plus to Spirit Shop 3
x4 = number supplied by Imprint-Sweats to Spirit Shop 1
x5 = number supplied by Imprint-Sweats to Spirit Shop 2
x6 = number supplied by Imprint-Sweats to Spirit Shop 3
The system of equations is
x1 + x4 = 15
x2 + x5 = 20
x3 + x6 = 30
x1 + x2 + x3 = 40
x4 + x5 + x6 = 25
The solution is
x1 = –10 + x5 + x6
x2 = 20 – x5
x3 = 30 – x6
x4 = 25 – x5 – x6
(b)x5 = 5, x6 = 5 yields x1 = 0, x2 = 15, x3 = 25, x4 = 15
x5 = 10, x6 = 5 yields x1 = 5, x2 = 10, x3 = 25, x4 = 10
x5 = 0, x6 = 15 yields x1 = 5, x2 = 20, x3 = 15, x4 = 10
(c)x5 represents the number of sweatshirts supplied by Imprint-Sweats to Spirit Shop 2. Since x2 = 20 – x5, x5 must be no greater than 20 or else x2 would be a negative number. Thus, x5 ² 20.
(d)If Imprint-Sweats supplies no sweatshirts to Spirit Shop 2 and Spirit Shop 3, then x5 = 0 and x6 = 0 so x1 = –10. Thus, the order cannot be filled.
(e)Since x1 = –10 + x5 + x6, in order for x1 ³ 0, then
x5 + x6 ³ 10. Imprint-Sweats must supply a total of 10 or more to Spirit Shops 2 and 3.
(f)Since x2 = 20 – x5, and 0 ² x5 ² 20, then x2 ² 20. Thus, Sweats-Plus supplies 20 or less to Spirit Shop 2.
74.The desired system is
x – 2y = 3
–2x + 4y = –6
75.The reduced echelon form of the augmented matrix is
which indicates no solution. In order to have a solution, the 2 in the third equation must be changed so that the bottom row of the reduced matrix is all zeros. Substitute c for 2 in the third equation and reduce the augmented matrix.
The augmented matrix is now which reduces to
We need only to reduce the matrix further by getting a zero in row 3, column 2.
Since c - 5 must be zero, c = 5. The last equation should be 3x + y + 4z = 5.
76.Reduce the augmented matrix.
so the last row is all zeros
Subtract row 2 from row 3
Since c- 1 = 0, c = 1. The last equation should be x + 4y – 3z = 1.
81.The system has infinitely many solutions.
82.(a)Let x1 = number of vehicles on Clay between A and B.
Let x2 = number of vehicles on 4th St. between B and C.
Let x3 = number of vehicles on Webster Ave. between C and D.
Let x4 = number of vehicles on 5th St. between A and D.
Since the number of vehicles entering an intersection equals the number leaving we have the following equations.
At A:x1 + x4 = 1200
At B:x1 + x2 = 1100
At C:x2 + x3 = 900
At D:x3 + x4 = 1000
(b)The solution is
x1 = 1200 – x4
x2 = –100 + x4
x3 = 1000 – x4
(c)Since traffic flow is nonnegative
x2 = –100 + x4 ³ 0. This implies x4 ³ 100.
Also x1 = 1200 – x4 ³ 0 and x3 = 1000 – x4 ³ 0.
In order for both to be true, x4 ² 1000. Thus, the minimum number of vehicles on 5th St. is 100 vehicles per hour and the maximum is 1000 vehicles per hour.
Using the minimum and maximum values of x4 in x1= 1200 – x4 we get that x1 ranges from 1100 to 200 so the minimum and maximum traffic flows on Clay Avenue are 200 and 1100 vehicles per hour, respectively.
Similarly, x3 ranges from 900 to 0 so the minimum and maximum traffic flows on Webster Ave. are 0 and 900 vehicles per hour, respectively.
(d)x1 = flow on Clay, x2 = flow on 4th Street, x3 = flow on Webster, x4 = flow on 5th Street
Then we have the following relationships.
At A: 800 + 400 - x4 = x1
At B: x1 + x2 = 500 + 600
At C: x2 + x3 = 300 + 600
At D: 700 + 300 = x3 + x4
We can write x2, x3, and x4 in terms of x1 as
x4 = 1200 - x1
x2 = 1100 - x1
x3 = 900 - x2 = 900 - (1100 - x1) = x1 - 200
For x4 to be nonnegative 1200 - x1 ³ 0 so x1 ² 1200
For x2 to be nonnegative 1100 - x1 ³ 0 so x1 ² 1100
For x3 to be nonnegative x1 - 200 ³ 0 so x1 ³ 200
For all three to be nonnegative 1100 ³ x1 ³ 200
The largest value of x1, the traffic flow on Clay Street, is 1100. In this case the traffic on 4th Street is at a standstill, the traffic on 5th Street is 100, and the traffic on Webster is heaviest, 900.
83. /(a)Label the traffic flow on each block by x1, x2, ..., x7 as shown. Then the condition that incoming traffic equals outgoing traffic at each intersection gives:
At A: x1 + x7 = 700
At B: x1 + x2 = 700
At C: x2 + x5 = 600 + x3
At D: x3 + 300 = 200 + x4
At E: 600 + x4 = 800 + x6
At F: x6 + 700 = x5 + x7
(b)The solution to the above system is
x1 = 700 – x7
x2 = x7
x3 = 100 + x6
x4 = 200 + x6
x5 = 700 + x6 – x7
(c)x3 represents the traffic flow on 10th St. between Colcord and Blair. Since x3 = 100 + x6, x3 ³ 100. The minimum traffic flow is 100 vehicles per hour when x6 = 0. The other traffic flows then become
x1 = 700 – x7
x2 = x7
x4 = 200
x5 = 700 – x7
84.Let x1 = number in first shift, x2 = number in second shift, x3 = number in third shift
x1 + x2 = 12(Lunch)
x2 + x3 = 8(Evening)
x1 - x3 = 5(Morning more than evening)
This system has no solution. A solution exists if 13 employees are used for the lunch peak.
85.c 14
86.(a)(b)
(c)(d)
(e)
87.x1 = 0.2 + 1.6x388.No solution
x2 = –0.2 + 1.4x3
89.(–19, 28, -10)90.No solution
91.(a)x1 = 0.5 + 3.5x4
x2 = 0.5 – 1.83x4
x3 = 2.33x4
(b)x2 = 0.5 – 1.83x4 implies 0.5 – 1.83x4 > 0
so x4 = 0.273.
Thus, 0 < x4 < 0.273
x3 = 2.33x4 implies 0 < x3 < 0.636
x1 = 0.5 + 3.5x4 implies x1 > 0.5 and the maximum value of x1 occurs when x4 is maximum. Thus, x1 < 0.5 + 3.5(0.273) = 1.456. 0.50 < x1 < 1.456
x2 = 0.5 – 1.83x4 implies x2 < 0.5 and x2 is minimum, 0, when x4 reaches 0.273 so
0 < x2 < 0.5.
92.Let x1 = number of T-shirts, x2 = number of white sweat shirts, and x3 = number of gold sweat shirts.
x1 + x2 + x3 = 1000(Total number)
7.80x1 + 16.50x2 + 18.00x3 = 11,505(Cost)
13.50x1 + 28.00x2 + 30.00x3 = 19,600(Revenue)
The augmented matrix for the system is
The Student Store bought 600 T-shirts, 250 white sweat shirts, and 150 gold sweat shirts.
Using Your TI-83
1.(2, -3, 4)2.(2.2, 1.5, 3.4)
3.(3, 4, 6)4.(1.1, 2.2, 3.3)
1
Section 2.4 Matrix Operations
1.2.(3, -2, 5)
3.(-2, 5, -3)4.(4, 2, 5)
Using Excel
1. (-5, 2, 6)2. (2, 7, 3)
3. (6, 2, 7)4. (4, 1, 2, 1)
Section 2.4
1. Alpha Beta2.Washers Dryers Micros
3. Joe Jane Judy4.Test 1 Test 2 Test 3
5.2 by 26.3 by 27.3 by 38.3 by 4
9.4 by 110.1 by 511.2 by 412.2 by 1
13.2 by 314.1 by 615.2 by 216.3 by 3
17.Not equal18.Equal 19.Equal
20.Not equal
21.Not equal 22.Not equal 23.24.
25.26.
27. Can not add them28. Can not add them
29.30.
31.32.
33.34.
35. 36.
37.
38.(a)A = B =
(b)A + B =
39.(a)3A = , -2B = , 5C =
(b)A + C =
(c)3A - 2B =
(d)A - 2B + 5C =
40.(a)A - 2B = - =
(b)-C = B - C =
(c)2A + 3B - 4C = + + =
41. I II III42.x = 9
43.x = 344.2x + 3 = 3x - 1 so x = 4
45.6x + 4 = 14x - 13 so x = 46.2x + 1 = 3x + 5 so x = -4
47.48.114 =
49. S.D. N.O. P.M.
12M =
50. 1 2
= =
51. = =
52.Class
FR SO JR SR
GPA
53.(a)Each row in a matrix represents a school, in the order given, and the columns represent students, tuition, and room and board in that order.
Y2002 =Y1994 =
The change in the data is found by subtracting:
Y2002 - Y1994 =
(b)Bowdoin had the largest increase in tuition, $7990.
(c)Maraquette had a decrease of 913 students, and Samford had a decrease of 324 students.
54.Each row represents a war in the order given and each column represents veterans, childlren, and spouses.
Decrease =
57.(a)(b)
(c)(d)
58.(a)(b)
(c)(d)
59.(a)(b)
(c) (d)
(e)(f)
Using Your TI- 83
1.(a)(b)
(c)(d)
2.(a)(b)
(c)(d)
Using EXCEL
1.2.
3.4.
5.
6.A + B = A - B = 4A =
0.35B = 4A + 6B =
1