1. Matrix Operations Given: A = [ 33 ]A-[3-321] -3 B = [342]B-[3-41-2] (a) A² A2 Multiply A× A: -3 = (3 x 32x-3) (3 x 22 x 1) | = |[19–63 |-9-3 -6+21] = A² = 33 33 1-3×3+1x-3) (-3×2+1x1) [12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1 ]=[3-128-5] (b) | A ||A| Determinant of A | A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9 (c) Adjoint of A Swap diagonal elements and change sign of off-diagonals: A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23] -3 (d) B-¹B-1 First find | B ||B|: |B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2 Then the adjoint of B: adj (B) = [² 3 adj(B)=[-24-13] Now, B-1 1 = |B| · adj (B) = 1 [²¯¯³¹³] = [2₂ B 0.5 |B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5] 2. (a) Matrix Method: Solve (2x + 3y = 6 (2x-3y=14 {2x+3y=62x-3y=14 Matrix form: 22 33-22 = [223-3][xy]=[614] Find inverse of coefficient matrix: Determinant: | M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12 Adjoint: adj(M) = [3]adj(M)-[-3-2-32] So inverse: M-1 = Multiply: 1 [323]-[0.167 0.25 -0.167]M-1 -1=-121[-3-2-32]=[0.250.1670.25-0.167] =
1. Matrix Operations Given: A = [ 33 ]A-[3-321] -3 B = [342]B-[3-41-2] (a) A² A2 Multiply A× A: -3 = (3 x 32x-3) (3 x 22 x 1) | = |[19–63 |-9-3 -6+21] = A² = 33 33 1-3×3+1x-3) (-3×2+1x1) [12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1 ]=[3-128-5] (b) | A ||A| Determinant of A | A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9 (c) Adjoint of A Swap diagonal elements and change sign of off-diagonals: A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23] -3 (d) B-¹B-1 First find | B ||B|: |B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2 Then the adjoint of B: adj (B) = [² 3 adj(B)=[-24-13] Now, B-1 1 = |B| · adj (B) = 1 [²¯¯³¹³] = [2₂ B 0.5 |B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5] 2. (a) Matrix Method: Solve (2x + 3y = 6 (2x-3y=14 {2x+3y=62x-3y=14 Matrix form: 22 33-22 = [223-3][xy]=[614] Find inverse of coefficient matrix: Determinant: | M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12 Adjoint: adj(M) = [3]adj(M)-[-3-2-32] So inverse: M-1 = Multiply: 1 [323]-[0.167 0.25 -0.167]M-1 -1=-121[-3-2-32]=[0.250.1670.25-0.167] =
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.3: Matrix Algebra
Problem 33E
Related questions
Question
![1. Matrix Operations
Given:
A = [ 33 ]A-[3-321]
-3
B = [342]B-[3-41-2]
(a) A² A2
Multiply A× A:
-3
=
(3 x 32x-3) (3 x 22 x 1)
| = |[19–63
|-9-3 -6+21] =
A² = 33 33 1-3×3+1x-3) (-3×2+1x1)
[12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1
]=[3-128-5]
(b) | A ||A| Determinant of A
| A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9
(c) Adjoint of A
Swap diagonal elements and change sign of off-diagonals:
A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23]
-3
(d) B-¹B-1
First find | B ||B|:
|B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2
Then the adjoint of B:
adj (B) = [²
3
adj(B)=[-24-13]
Now,
B-1
1
=
|B|
· adj (B) = 1 [²¯¯³¹³] = [2₂ B
0.5
|B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5]
2.
(a) Matrix Method: Solve
(2x + 3y = 6
(2x-3y=14
{2x+3y=62x-3y=14
Matrix form:
22 33-22
=
[223-3][xy]=[614]
Find inverse of coefficient matrix: Determinant:
| M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12
Adjoint:
adj(M) = [3]adj(M)-[-3-2-32]
So inverse:
M-1
=
Multiply:
1
[323]-[0.167
0.25
-0.167]M-1
-1=-121[-3-2-32]=[0.250.1670.25-0.167]
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F72d747fa-6860-4cf8-a406-b6f67bb39e6b%2F7a2ee242-a94b-4153-8c30-1d92a83d96e9%2Fn9evnzo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Matrix Operations
Given:
A = [ 33 ]A-[3-321]
-3
B = [342]B-[3-41-2]
(a) A² A2
Multiply A× A:
-3
=
(3 x 32x-3) (3 x 22 x 1)
| = |[19–63
|-9-3 -6+21] =
A² = 33 33 1-3×3+1x-3) (-3×2+1x1)
[12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1
]=[3-128-5]
(b) | A ||A| Determinant of A
| A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9
(c) Adjoint of A
Swap diagonal elements and change sign of off-diagonals:
A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23]
-3
(d) B-¹B-1
First find | B ||B|:
|B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2
Then the adjoint of B:
adj (B) = [²
3
adj(B)=[-24-13]
Now,
B-1
1
=
|B|
· adj (B) = 1 [²¯¯³¹³] = [2₂ B
0.5
|B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5]
2.
(a) Matrix Method: Solve
(2x + 3y = 6
(2x-3y=14
{2x+3y=62x-3y=14
Matrix form:
22 33-22
=
[223-3][xy]=[614]
Find inverse of coefficient matrix: Determinant:
| M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12
Adjoint:
adj(M) = [3]adj(M)-[-3-2-32]
So inverse:
M-1
=
Multiply:
1
[323]-[0.167
0.25
-0.167]M-1
-1=-121[-3-2-32]=[0.250.1670.25-0.167]
=
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