Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section: Chapter Questions
Problem 59RE
Related questions
Question
![VoWiFi 1
Exercise Set 2.3
► In Exercises 1-4, verify that det (kA) = k" det(A). ◄
1. A =
2. A = [3_-3]; k=
-2
5. A =
1
3
4
2 -1 3
3. A = 3 2 1: k=-2
dit
1 4 5
1 1 1
4. A = 0 2 3 k 3
0
1
6. A =
7. A =
► In Exercises 5-6, verify that det (AB) = det (BA) and deter-
mine whether the equality det(A + B) = det (A) + det (B) holds.
9. A =
11. A =
13. A =
15. A =
Properties of De...
1
-2
17. A =
; k= 2
2 1 0
3 4 0 and B
0 0
2
2
0
0
► In Exercises 7-14, use determinants to decide whether the given
matrix is invertible. <
4
-2
3
2 5
-1 -1
2 4
8 2
0 -1
2 2
2 0
8 1
-5 3
-3
1 -3
0 2
2.2
K/s
00
Q3%
20:17
it suffices to show that each minor Mi¡ with i < j is zero. For this purpose, let B¡¡ b
matrix that results when the ith row and jth column of A are deleted, so
2
1
1
0
6
Mij
Q
From the assumption that i <j, it follows Bij is upper angular (see Figure 1
Since A is upper triangular, its (i + 1)-st row begins with at least i zeros. But the it
of Dij is the (i | 1) st row of A with the entry in the jth column removed. Sinee
none of the first i zeros is removed by deleting the jth column; thus the ith row
starts with at least i zeros, which implies that this row has a zero on the main diag
It now follows from Theorem 2.1.2 that det (Bij) = 0 and from (10) that Mij = 0.
1
2
4
3 1 6
k 3 2
0
3
-4
[k - 3 -2
-2 k-2
1 -1 3
7
1
2
5
0
and B =
8. A =
10. A =
12. A =
0
16. A =
NON
18. A =
0 3
0 3 2
-2 0 -4
-3 0
►In Exercises 15-18, find the values of k for which the matrix A
is invertible. ◄
0 6
5
8 0 3
√2 -√7
14. A = 3√2-3√7
5
-4
1
0
9 -1 4
9
1
2
k
1
k
0 2 1
0
0
-9 0
2.3 Properties of Determinants; Cramer's
► In Exercises 19-23, decide whether the matrix is inv
if so, use the adjoint method to find its inverse.
19. A =
21. A =
23. A =
Bij)
28.
-1
2
0
0
2 5 5
-1
0
2 4
3
-3
1
0
1
2
1
3
2
5
1
3
1 3 22
5
-3
2
1
2
26. x 4y + z =
8 9
NGON
6
-1
4x = y + 2z =
2x + 2y3z = -20
20. A =
29. 3x₁ - x₂ + x3 = 4
-x₁ +7x₂2x3 = 1
2x₁ + 6x₂x3 = 5
30. Show that the matrix
22. A =
A =
► In Exercises 24-29, solve by Cramer's rule, where
24. 7x₁2x₂ = 3
25. 4x + 5y
3x₁ + x₂ = 5
x₁4x₂ + 2x3 + x4 = -32
2x₁x₂7x3 + 9x4 = 14
11
-x₁ + x₂ + 3x3 + x4 =
x₁2x₂ + x3 4x4 =
-4
2
0
-2
cos
- sin 0
0
2
0
8 1
3
-5
sin 0
cos
0
27. x₁3x2 + x3
2x1x₂
4x1
0
3
0
11x + y + 2z =
x + 5y + 2z =
0
C
C
32. Let Ax=b be the system in Exercise 31.
(a) Solve by Cramer's rule.
(b) Solve by Gauss-Jord
(c) Which method invo
=
4x+y+z+ w= 6
3x + 7yz + w= 1
7x + 3y5z +8w=-3
x + y + z + 2w= 3
- 3x3
1
is invertible for all values of 0; then find A-¹ u
rem 2.3.6.
31. Use Cramer's rule to solve for y without solving
knowns x, z, and w.
ions?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9085df12-c476-42f7-81b7-8c22449f1354%2F5a370249-6899-407c-9e14-80fb5cad6896%2Fe4l4038_processed.jpeg&w=3840&q=75)
Transcribed Image Text:VoWiFi 1
Exercise Set 2.3
► In Exercises 1-4, verify that det (kA) = k" det(A). ◄
1. A =
2. A = [3_-3]; k=
-2
5. A =
1
3
4
2 -1 3
3. A = 3 2 1: k=-2
dit
1 4 5
1 1 1
4. A = 0 2 3 k 3
0
1
6. A =
7. A =
► In Exercises 5-6, verify that det (AB) = det (BA) and deter-
mine whether the equality det(A + B) = det (A) + det (B) holds.
9. A =
11. A =
13. A =
15. A =
Properties of De...
1
-2
17. A =
; k= 2
2 1 0
3 4 0 and B
0 0
2
2
0
0
► In Exercises 7-14, use determinants to decide whether the given
matrix is invertible. <
4
-2
3
2 5
-1 -1
2 4
8 2
0 -1
2 2
2 0
8 1
-5 3
-3
1 -3
0 2
2.2
K/s
00
Q3%
20:17
it suffices to show that each minor Mi¡ with i < j is zero. For this purpose, let B¡¡ b
matrix that results when the ith row and jth column of A are deleted, so
2
1
1
0
6
Mij
Q
From the assumption that i <j, it follows Bij is upper angular (see Figure 1
Since A is upper triangular, its (i + 1)-st row begins with at least i zeros. But the it
of Dij is the (i | 1) st row of A with the entry in the jth column removed. Sinee
none of the first i zeros is removed by deleting the jth column; thus the ith row
starts with at least i zeros, which implies that this row has a zero on the main diag
It now follows from Theorem 2.1.2 that det (Bij) = 0 and from (10) that Mij = 0.
1
2
4
3 1 6
k 3 2
0
3
-4
[k - 3 -2
-2 k-2
1 -1 3
7
1
2
5
0
and B =
8. A =
10. A =
12. A =
0
16. A =
NON
18. A =
0 3
0 3 2
-2 0 -4
-3 0
►In Exercises 15-18, find the values of k for which the matrix A
is invertible. ◄
0 6
5
8 0 3
√2 -√7
14. A = 3√2-3√7
5
-4
1
0
9 -1 4
9
1
2
k
1
k
0 2 1
0
0
-9 0
2.3 Properties of Determinants; Cramer's
► In Exercises 19-23, decide whether the matrix is inv
if so, use the adjoint method to find its inverse.
19. A =
21. A =
23. A =
Bij)
28.
-1
2
0
0
2 5 5
-1
0
2 4
3
-3
1
0
1
2
1
3
2
5
1
3
1 3 22
5
-3
2
1
2
26. x 4y + z =
8 9
NGON
6
-1
4x = y + 2z =
2x + 2y3z = -20
20. A =
29. 3x₁ - x₂ + x3 = 4
-x₁ +7x₂2x3 = 1
2x₁ + 6x₂x3 = 5
30. Show that the matrix
22. A =
A =
► In Exercises 24-29, solve by Cramer's rule, where
24. 7x₁2x₂ = 3
25. 4x + 5y
3x₁ + x₂ = 5
x₁4x₂ + 2x3 + x4 = -32
2x₁x₂7x3 + 9x4 = 14
11
-x₁ + x₂ + 3x3 + x4 =
x₁2x₂ + x3 4x4 =
-4
2
0
-2
cos
- sin 0
0
2
0
8 1
3
-5
sin 0
cos
0
27. x₁3x2 + x3
2x1x₂
4x1
0
3
0
11x + y + 2z =
x + 5y + 2z =
0
C
C
32. Let Ax=b be the system in Exercise 31.
(a) Solve by Cramer's rule.
(b) Solve by Gauss-Jord
(c) Which method invo
=
4x+y+z+ w= 6
3x + 7yz + w= 1
7x + 3y5z +8w=-3
x + y + z + 2w= 3
- 3x3
1
is invertible for all values of 0; then find A-¹ u
rem 2.3.6.
31. Use Cramer's rule to solve for y without solving
knowns x, z, and w.
ions?
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